php retrieve data from mySQL database into form












0















I have a registration form, and what I like to so is to use a mySQL query to retrieve data in to the form input fields.



this is part of the very long form:



<form class="form-horizontal" id="registration" method='post' action='ConnectDB.php' onsubmit='return ValidateForm(this);'>
<fieldset>
<h1 style="margin-top: px;float:right; color:#62A1D5"><br><b>registration</b></h1>
<table id="TBackGround">
<td style="padding-left:10px; padding-right:10px;">
<p id="pd" style="margin-top:5px; margin-right:2px; font-size:16px; text-decoration:underline"><b><br>details:</b><p>
<p style="line-height: 1.3em" class="details">
<div class="control-group">
<label class="control-label">ID </label>
<input type="text"name="studentID" align= "right" class="textbox" ><span id="errorID"></span> <br/>
</div>
<div class="control-group">
<label class="control-label" >First name</label>
<input type="text" name="Fname" class="textbox" ><span id="errorFirstName"> <br/>
</div>
</form>


How do I set the retrived data to be loaded in to the form's input fields?
What I have is a query to retrieve the ID of the record but I don't know how to set the entire query result on to the fields.



my php query:



<?php 

if(isset($_POST['submit']))
{
$query = $_POST['query'];
$min_length = 1;
if(strlen($query) >= $min_length)
{
$query = htmlspecialchars($query);
$query = mysql_real_escape_string($query);
echo "<table border='0' width='300' align='center' cellpadding='1' cellspacing='1'>";
echo "<tr align='center' bgcolor='#002C40'>
?>
<td height='35px' width='150px'>id</td> <td>first name</td>

</tr>";
$raw_results =

mysql_query("SELECT * FROM student WHERE (`idStudent` LIKE '%".$query."%') OR (`FirstName` LIKE '%".$query."%')");
if(mysql_num_rows($raw_results) > 0)
{
while($results = mysql_fetch_array($raw_results))
{

echo "<tr align='center' bgcolor='#0f7ea3'>

<td height='25px'> "
.$results['idStudent']."</td> <td>".$results['FirstName']."</td>






</tr>" ;
}

}
else{
echo "<tr align='center' bgcolor='#6C0000'>

<td colspan='2' height='25px'>No results</td><tr>";
echo "</table>";
}
}
else{
echo "Minimum length is ".$min_length;
}}

?>









share|improve this question

























  • you mean: suggestion, auto-complete? Than you need AJAX.

    – raiserle
    Apr 16 '14 at 14:29











  • Can I get an example, even a very simple one?

    – user2674835
    Apr 16 '14 at 14:30











  • @user2674835: what you mean? Example of suggestion, auto-complete?

    – raiserle
    Apr 16 '14 at 14:34











  • an example of using a simple select query and loading the data in to input fields

    – user2674835
    Apr 16 '14 at 14:37








  • 1





    it would be very helpful if you included only the relevant code portions

    – bsapaka
    Apr 16 '14 at 14:40
















0















I have a registration form, and what I like to so is to use a mySQL query to retrieve data in to the form input fields.



this is part of the very long form:



<form class="form-horizontal" id="registration" method='post' action='ConnectDB.php' onsubmit='return ValidateForm(this);'>
<fieldset>
<h1 style="margin-top: px;float:right; color:#62A1D5"><br><b>registration</b></h1>
<table id="TBackGround">
<td style="padding-left:10px; padding-right:10px;">
<p id="pd" style="margin-top:5px; margin-right:2px; font-size:16px; text-decoration:underline"><b><br>details:</b><p>
<p style="line-height: 1.3em" class="details">
<div class="control-group">
<label class="control-label">ID </label>
<input type="text"name="studentID" align= "right" class="textbox" ><span id="errorID"></span> <br/>
</div>
<div class="control-group">
<label class="control-label" >First name</label>
<input type="text" name="Fname" class="textbox" ><span id="errorFirstName"> <br/>
</div>
</form>


How do I set the retrived data to be loaded in to the form's input fields?
What I have is a query to retrieve the ID of the record but I don't know how to set the entire query result on to the fields.



my php query:



<?php 

if(isset($_POST['submit']))
{
$query = $_POST['query'];
$min_length = 1;
if(strlen($query) >= $min_length)
{
$query = htmlspecialchars($query);
$query = mysql_real_escape_string($query);
echo "<table border='0' width='300' align='center' cellpadding='1' cellspacing='1'>";
echo "<tr align='center' bgcolor='#002C40'>
?>
<td height='35px' width='150px'>id</td> <td>first name</td>

</tr>";
$raw_results =

mysql_query("SELECT * FROM student WHERE (`idStudent` LIKE '%".$query."%') OR (`FirstName` LIKE '%".$query."%')");
if(mysql_num_rows($raw_results) > 0)
{
while($results = mysql_fetch_array($raw_results))
{

echo "<tr align='center' bgcolor='#0f7ea3'>

<td height='25px'> "
.$results['idStudent']."</td> <td>".$results['FirstName']."</td>






</tr>" ;
}

}
else{
echo "<tr align='center' bgcolor='#6C0000'>

<td colspan='2' height='25px'>No results</td><tr>";
echo "</table>";
}
}
else{
echo "Minimum length is ".$min_length;
}}

?>









share|improve this question

























  • you mean: suggestion, auto-complete? Than you need AJAX.

    – raiserle
    Apr 16 '14 at 14:29











  • Can I get an example, even a very simple one?

    – user2674835
    Apr 16 '14 at 14:30











  • @user2674835: what you mean? Example of suggestion, auto-complete?

    – raiserle
    Apr 16 '14 at 14:34











  • an example of using a simple select query and loading the data in to input fields

    – user2674835
    Apr 16 '14 at 14:37








  • 1





    it would be very helpful if you included only the relevant code portions

    – bsapaka
    Apr 16 '14 at 14:40














0












0








0








I have a registration form, and what I like to so is to use a mySQL query to retrieve data in to the form input fields.



this is part of the very long form:



<form class="form-horizontal" id="registration" method='post' action='ConnectDB.php' onsubmit='return ValidateForm(this);'>
<fieldset>
<h1 style="margin-top: px;float:right; color:#62A1D5"><br><b>registration</b></h1>
<table id="TBackGround">
<td style="padding-left:10px; padding-right:10px;">
<p id="pd" style="margin-top:5px; margin-right:2px; font-size:16px; text-decoration:underline"><b><br>details:</b><p>
<p style="line-height: 1.3em" class="details">
<div class="control-group">
<label class="control-label">ID </label>
<input type="text"name="studentID" align= "right" class="textbox" ><span id="errorID"></span> <br/>
</div>
<div class="control-group">
<label class="control-label" >First name</label>
<input type="text" name="Fname" class="textbox" ><span id="errorFirstName"> <br/>
</div>
</form>


How do I set the retrived data to be loaded in to the form's input fields?
What I have is a query to retrieve the ID of the record but I don't know how to set the entire query result on to the fields.



my php query:



<?php 

if(isset($_POST['submit']))
{
$query = $_POST['query'];
$min_length = 1;
if(strlen($query) >= $min_length)
{
$query = htmlspecialchars($query);
$query = mysql_real_escape_string($query);
echo "<table border='0' width='300' align='center' cellpadding='1' cellspacing='1'>";
echo "<tr align='center' bgcolor='#002C40'>
?>
<td height='35px' width='150px'>id</td> <td>first name</td>

</tr>";
$raw_results =

mysql_query("SELECT * FROM student WHERE (`idStudent` LIKE '%".$query."%') OR (`FirstName` LIKE '%".$query."%')");
if(mysql_num_rows($raw_results) > 0)
{
while($results = mysql_fetch_array($raw_results))
{

echo "<tr align='center' bgcolor='#0f7ea3'>

<td height='25px'> "
.$results['idStudent']."</td> <td>".$results['FirstName']."</td>






</tr>" ;
}

}
else{
echo "<tr align='center' bgcolor='#6C0000'>

<td colspan='2' height='25px'>No results</td><tr>";
echo "</table>";
}
}
else{
echo "Minimum length is ".$min_length;
}}

?>









share|improve this question
















I have a registration form, and what I like to so is to use a mySQL query to retrieve data in to the form input fields.



this is part of the very long form:



<form class="form-horizontal" id="registration" method='post' action='ConnectDB.php' onsubmit='return ValidateForm(this);'>
<fieldset>
<h1 style="margin-top: px;float:right; color:#62A1D5"><br><b>registration</b></h1>
<table id="TBackGround">
<td style="padding-left:10px; padding-right:10px;">
<p id="pd" style="margin-top:5px; margin-right:2px; font-size:16px; text-decoration:underline"><b><br>details:</b><p>
<p style="line-height: 1.3em" class="details">
<div class="control-group">
<label class="control-label">ID </label>
<input type="text"name="studentID" align= "right" class="textbox" ><span id="errorID"></span> <br/>
</div>
<div class="control-group">
<label class="control-label" >First name</label>
<input type="text" name="Fname" class="textbox" ><span id="errorFirstName"> <br/>
</div>
</form>


How do I set the retrived data to be loaded in to the form's input fields?
What I have is a query to retrieve the ID of the record but I don't know how to set the entire query result on to the fields.



my php query:



<?php 

if(isset($_POST['submit']))
{
$query = $_POST['query'];
$min_length = 1;
if(strlen($query) >= $min_length)
{
$query = htmlspecialchars($query);
$query = mysql_real_escape_string($query);
echo "<table border='0' width='300' align='center' cellpadding='1' cellspacing='1'>";
echo "<tr align='center' bgcolor='#002C40'>
?>
<td height='35px' width='150px'>id</td> <td>first name</td>

</tr>";
$raw_results =

mysql_query("SELECT * FROM student WHERE (`idStudent` LIKE '%".$query."%') OR (`FirstName` LIKE '%".$query."%')");
if(mysql_num_rows($raw_results) > 0)
{
while($results = mysql_fetch_array($raw_results))
{

echo "<tr align='center' bgcolor='#0f7ea3'>

<td height='25px'> "
.$results['idStudent']."</td> <td>".$results['FirstName']."</td>






</tr>" ;
}

}
else{
echo "<tr align='center' bgcolor='#6C0000'>

<td colspan='2' height='25px'>No results</td><tr>";
echo "</table>";
}
}
else{
echo "Minimum length is ".$min_length;
}}

?>






php html mysql html5






share|improve this question















share|improve this question













share|improve this question




share|improve this question








edited Apr 16 '14 at 14:24









Reinder Wit

5,39311628




5,39311628










asked Apr 16 '14 at 14:14









user2674835user2674835

641310




641310













  • you mean: suggestion, auto-complete? Than you need AJAX.

    – raiserle
    Apr 16 '14 at 14:29











  • Can I get an example, even a very simple one?

    – user2674835
    Apr 16 '14 at 14:30











  • @user2674835: what you mean? Example of suggestion, auto-complete?

    – raiserle
    Apr 16 '14 at 14:34











  • an example of using a simple select query and loading the data in to input fields

    – user2674835
    Apr 16 '14 at 14:37








  • 1





    it would be very helpful if you included only the relevant code portions

    – bsapaka
    Apr 16 '14 at 14:40



















  • you mean: suggestion, auto-complete? Than you need AJAX.

    – raiserle
    Apr 16 '14 at 14:29











  • Can I get an example, even a very simple one?

    – user2674835
    Apr 16 '14 at 14:30











  • @user2674835: what you mean? Example of suggestion, auto-complete?

    – raiserle
    Apr 16 '14 at 14:34











  • an example of using a simple select query and loading the data in to input fields

    – user2674835
    Apr 16 '14 at 14:37








  • 1





    it would be very helpful if you included only the relevant code portions

    – bsapaka
    Apr 16 '14 at 14:40

















you mean: suggestion, auto-complete? Than you need AJAX.

– raiserle
Apr 16 '14 at 14:29





you mean: suggestion, auto-complete? Than you need AJAX.

– raiserle
Apr 16 '14 at 14:29













Can I get an example, even a very simple one?

– user2674835
Apr 16 '14 at 14:30





Can I get an example, even a very simple one?

– user2674835
Apr 16 '14 at 14:30













@user2674835: what you mean? Example of suggestion, auto-complete?

– raiserle
Apr 16 '14 at 14:34





@user2674835: what you mean? Example of suggestion, auto-complete?

– raiserle
Apr 16 '14 at 14:34













an example of using a simple select query and loading the data in to input fields

– user2674835
Apr 16 '14 at 14:37







an example of using a simple select query and loading the data in to input fields

– user2674835
Apr 16 '14 at 14:37






1




1





it would be very helpful if you included only the relevant code portions

– bsapaka
Apr 16 '14 at 14:40





it would be very helpful if you included only the relevant code portions

– bsapaka
Apr 16 '14 at 14:40












2 Answers
2






active

oldest

votes


















0














OK, Basics!



You've an form with input-elements. By sending the form to the server, the server-side-script can get the values of the form-elements.



Look this:



<form action="the_script_location.php" method="post">
<input type="text" name="element1" />
<input type="submit" name="send" />
</form>


On the server-side-script you can now get the values from form like this:



<?php
if( isset( $_POST[ 'send' ] ) ){
$var = $_POST[ 'element1' ];
//now you can use the var for an query to your database
//please note: this very basic, without any security of injection
$res = mysql_query( 'SELECT `any` FROM `anyTable` WHERE `any` LIKE '%'.$var.'%'' );
if( mysql_num_row( $res ){
$row = mysql_fetch_assoc( $res ); //get one (first) result
}
}
?>


Now you can update the form:



<form action="the_script_location.php" method="post">
<input type="text" name="element1" value="<?php isset( $row[ 'your_filed_in_database' ] ) ? $row[ 'your_filed_in_database' ] : '' ?>" />
<input type="submit" name="send" />
</form>





share|improve this answer
























  • Thanks for the example! However, I'm still stuck on understanding how to pass the query results to my form. shouldn't be a code for opening my desired page after the php code is done?

    – user2674835
    Apr 16 '14 at 16:00



















0














You need to output layout of your form from php file.
Here is small example can help you.



File register.php



if (!isset($_POST['submit'])) {
mysql_connect();
$res = mysql_query("SELECT * FROM users WHERE user_id=1");
$user = array_map('htmlspecialchars', mysql_fetch_assoc($res));

echo <<<CUT
<form action="register.php" method="post" >
<lable>Name:</label> <input type="text" name="name" value="{$user['name']}" />
<lable>Country:</label> <input type="text" name="name" value="{$user['country']}" />
<input type="submit" name="submit" value="Submit" />
</form>
CUT;
} else {
//handle submited data here
}





share|improve this answer























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    2 Answers
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    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    0














    OK, Basics!



    You've an form with input-elements. By sending the form to the server, the server-side-script can get the values of the form-elements.



    Look this:



    <form action="the_script_location.php" method="post">
    <input type="text" name="element1" />
    <input type="submit" name="send" />
    </form>


    On the server-side-script you can now get the values from form like this:



    <?php
    if( isset( $_POST[ 'send' ] ) ){
    $var = $_POST[ 'element1' ];
    //now you can use the var for an query to your database
    //please note: this very basic, without any security of injection
    $res = mysql_query( 'SELECT `any` FROM `anyTable` WHERE `any` LIKE '%'.$var.'%'' );
    if( mysql_num_row( $res ){
    $row = mysql_fetch_assoc( $res ); //get one (first) result
    }
    }
    ?>


    Now you can update the form:



    <form action="the_script_location.php" method="post">
    <input type="text" name="element1" value="<?php isset( $row[ 'your_filed_in_database' ] ) ? $row[ 'your_filed_in_database' ] : '' ?>" />
    <input type="submit" name="send" />
    </form>





    share|improve this answer
























    • Thanks for the example! However, I'm still stuck on understanding how to pass the query results to my form. shouldn't be a code for opening my desired page after the php code is done?

      – user2674835
      Apr 16 '14 at 16:00
















    0














    OK, Basics!



    You've an form with input-elements. By sending the form to the server, the server-side-script can get the values of the form-elements.



    Look this:



    <form action="the_script_location.php" method="post">
    <input type="text" name="element1" />
    <input type="submit" name="send" />
    </form>


    On the server-side-script you can now get the values from form like this:



    <?php
    if( isset( $_POST[ 'send' ] ) ){
    $var = $_POST[ 'element1' ];
    //now you can use the var for an query to your database
    //please note: this very basic, without any security of injection
    $res = mysql_query( 'SELECT `any` FROM `anyTable` WHERE `any` LIKE '%'.$var.'%'' );
    if( mysql_num_row( $res ){
    $row = mysql_fetch_assoc( $res ); //get one (first) result
    }
    }
    ?>


    Now you can update the form:



    <form action="the_script_location.php" method="post">
    <input type="text" name="element1" value="<?php isset( $row[ 'your_filed_in_database' ] ) ? $row[ 'your_filed_in_database' ] : '' ?>" />
    <input type="submit" name="send" />
    </form>





    share|improve this answer
























    • Thanks for the example! However, I'm still stuck on understanding how to pass the query results to my form. shouldn't be a code for opening my desired page after the php code is done?

      – user2674835
      Apr 16 '14 at 16:00














    0












    0








    0







    OK, Basics!



    You've an form with input-elements. By sending the form to the server, the server-side-script can get the values of the form-elements.



    Look this:



    <form action="the_script_location.php" method="post">
    <input type="text" name="element1" />
    <input type="submit" name="send" />
    </form>


    On the server-side-script you can now get the values from form like this:



    <?php
    if( isset( $_POST[ 'send' ] ) ){
    $var = $_POST[ 'element1' ];
    //now you can use the var for an query to your database
    //please note: this very basic, without any security of injection
    $res = mysql_query( 'SELECT `any` FROM `anyTable` WHERE `any` LIKE '%'.$var.'%'' );
    if( mysql_num_row( $res ){
    $row = mysql_fetch_assoc( $res ); //get one (first) result
    }
    }
    ?>


    Now you can update the form:



    <form action="the_script_location.php" method="post">
    <input type="text" name="element1" value="<?php isset( $row[ 'your_filed_in_database' ] ) ? $row[ 'your_filed_in_database' ] : '' ?>" />
    <input type="submit" name="send" />
    </form>





    share|improve this answer













    OK, Basics!



    You've an form with input-elements. By sending the form to the server, the server-side-script can get the values of the form-elements.



    Look this:



    <form action="the_script_location.php" method="post">
    <input type="text" name="element1" />
    <input type="submit" name="send" />
    </form>


    On the server-side-script you can now get the values from form like this:



    <?php
    if( isset( $_POST[ 'send' ] ) ){
    $var = $_POST[ 'element1' ];
    //now you can use the var for an query to your database
    //please note: this very basic, without any security of injection
    $res = mysql_query( 'SELECT `any` FROM `anyTable` WHERE `any` LIKE '%'.$var.'%'' );
    if( mysql_num_row( $res ){
    $row = mysql_fetch_assoc( $res ); //get one (first) result
    }
    }
    ?>


    Now you can update the form:



    <form action="the_script_location.php" method="post">
    <input type="text" name="element1" value="<?php isset( $row[ 'your_filed_in_database' ] ) ? $row[ 'your_filed_in_database' ] : '' ?>" />
    <input type="submit" name="send" />
    </form>






    share|improve this answer












    share|improve this answer



    share|improve this answer










    answered Apr 16 '14 at 14:59









    raiserleraiserle

    424420




    424420













    • Thanks for the example! However, I'm still stuck on understanding how to pass the query results to my form. shouldn't be a code for opening my desired page after the php code is done?

      – user2674835
      Apr 16 '14 at 16:00



















    • Thanks for the example! However, I'm still stuck on understanding how to pass the query results to my form. shouldn't be a code for opening my desired page after the php code is done?

      – user2674835
      Apr 16 '14 at 16:00

















    Thanks for the example! However, I'm still stuck on understanding how to pass the query results to my form. shouldn't be a code for opening my desired page after the php code is done?

    – user2674835
    Apr 16 '14 at 16:00





    Thanks for the example! However, I'm still stuck on understanding how to pass the query results to my form. shouldn't be a code for opening my desired page after the php code is done?

    – user2674835
    Apr 16 '14 at 16:00













    0














    You need to output layout of your form from php file.
    Here is small example can help you.



    File register.php



    if (!isset($_POST['submit'])) {
    mysql_connect();
    $res = mysql_query("SELECT * FROM users WHERE user_id=1");
    $user = array_map('htmlspecialchars', mysql_fetch_assoc($res));

    echo <<<CUT
    <form action="register.php" method="post" >
    <lable>Name:</label> <input type="text" name="name" value="{$user['name']}" />
    <lable>Country:</label> <input type="text" name="name" value="{$user['country']}" />
    <input type="submit" name="submit" value="Submit" />
    </form>
    CUT;
    } else {
    //handle submited data here
    }





    share|improve this answer




























      0














      You need to output layout of your form from php file.
      Here is small example can help you.



      File register.php



      if (!isset($_POST['submit'])) {
      mysql_connect();
      $res = mysql_query("SELECT * FROM users WHERE user_id=1");
      $user = array_map('htmlspecialchars', mysql_fetch_assoc($res));

      echo <<<CUT
      <form action="register.php" method="post" >
      <lable>Name:</label> <input type="text" name="name" value="{$user['name']}" />
      <lable>Country:</label> <input type="text" name="name" value="{$user['country']}" />
      <input type="submit" name="submit" value="Submit" />
      </form>
      CUT;
      } else {
      //handle submited data here
      }





      share|improve this answer


























        0












        0








        0







        You need to output layout of your form from php file.
        Here is small example can help you.



        File register.php



        if (!isset($_POST['submit'])) {
        mysql_connect();
        $res = mysql_query("SELECT * FROM users WHERE user_id=1");
        $user = array_map('htmlspecialchars', mysql_fetch_assoc($res));

        echo <<<CUT
        <form action="register.php" method="post" >
        <lable>Name:</label> <input type="text" name="name" value="{$user['name']}" />
        <lable>Country:</label> <input type="text" name="name" value="{$user['country']}" />
        <input type="submit" name="submit" value="Submit" />
        </form>
        CUT;
        } else {
        //handle submited data here
        }





        share|improve this answer













        You need to output layout of your form from php file.
        Here is small example can help you.



        File register.php



        if (!isset($_POST['submit'])) {
        mysql_connect();
        $res = mysql_query("SELECT * FROM users WHERE user_id=1");
        $user = array_map('htmlspecialchars', mysql_fetch_assoc($res));

        echo <<<CUT
        <form action="register.php" method="post" >
        <lable>Name:</label> <input type="text" name="name" value="{$user['name']}" />
        <lable>Country:</label> <input type="text" name="name" value="{$user['country']}" />
        <input type="submit" name="submit" value="Submit" />
        </form>
        CUT;
        } else {
        //handle submited data here
        }






        share|improve this answer












        share|improve this answer



        share|improve this answer










        answered Nov 12 '16 at 11:48









        Andrey YerokhinAndrey Yerokhin

        21016




        21016






























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