php retrieve data from mySQL database into form
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 ".$min_length;
}}
?>
php html mysql html5
add a comment |
I have a registration form, and what I like to so is to use a mySQL query to retrieve data in to the form input fields.
this is part of the very long form:
<form class="form-horizontal" id="registration" method='post' action='ConnectDB.php' onsubmit='return ValidateForm(this);'>
<fieldset>
<h1 style="margin-top: px;float:right; color:#62A1D5"><br><b>registration</b></h1>
<table id="TBackGround">
<td style="padding-left:10px; padding-right:10px;">
<p id="pd" style="margin-top:5px; margin-right:2px; font-size:16px; text-decoration:underline"><b><br>details:</b><p>
<p style="line-height: 1.3em" class="details">
<div class="control-group">
<label class="control-label">ID </label>
<input type="text"name="studentID" align= "right" class="textbox" ><span id="errorID"></span> <br/>
</div>
<div class="control-group">
<label class="control-label" >First name</label>
<input type="text" name="Fname" class="textbox" ><span id="errorFirstName"> <br/>
</div>
</form>
How do I set the retrived data to be loaded in to the form's input fields?
What I have is a query to retrieve the ID of the record but I don't know how to set the entire query result on to the fields.
my php query:
<?php
if(isset($_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 ".$min_length;
}}
?>
php html mysql html5
you mean: suggestion, auto-complete? Than you need AJAX.
– raiserle
Apr 16 '14 at 14:29
Can I get an example, even a very simple one?
– user2674835
Apr 16 '14 at 14:30
@user2674835: what you mean? Example of suggestion, auto-complete?
– raiserle
Apr 16 '14 at 14:34
an example of using a simple select query and loading the data in to input fields
– user2674835
Apr 16 '14 at 14:37
1
it would be very helpful if you included only the relevant code portions
– bsapaka
Apr 16 '14 at 14:40
add a comment |
I have a registration form, and what I like to so is to use a mySQL query to retrieve data in to the form input fields.
this is part of the very long form:
<form class="form-horizontal" id="registration" method='post' action='ConnectDB.php' onsubmit='return ValidateForm(this);'>
<fieldset>
<h1 style="margin-top: px;float:right; color:#62A1D5"><br><b>registration</b></h1>
<table id="TBackGround">
<td style="padding-left:10px; padding-right:10px;">
<p id="pd" style="margin-top:5px; margin-right:2px; font-size:16px; text-decoration:underline"><b><br>details:</b><p>
<p style="line-height: 1.3em" class="details">
<div class="control-group">
<label class="control-label">ID </label>
<input type="text"name="studentID" align= "right" class="textbox" ><span id="errorID"></span> <br/>
</div>
<div class="control-group">
<label class="control-label" >First name</label>
<input type="text" name="Fname" class="textbox" ><span id="errorFirstName"> <br/>
</div>
</form>
How do I set the retrived data to be loaded in to the form's input fields?
What I have is a query to retrieve the ID of the record but I don't know how to set the entire query result on to the fields.
my php query:
<?php
if(isset($_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 ".$min_length;
}}
?>
php html mysql html5
I have a registration form, and what I like to so is to use a mySQL query to retrieve data in to the form input fields.
this is part of the very long form:
<form class="form-horizontal" id="registration" method='post' action='ConnectDB.php' onsubmit='return ValidateForm(this);'>
<fieldset>
<h1 style="margin-top: px;float:right; color:#62A1D5"><br><b>registration</b></h1>
<table id="TBackGround">
<td style="padding-left:10px; padding-right:10px;">
<p id="pd" style="margin-top:5px; margin-right:2px; font-size:16px; text-decoration:underline"><b><br>details:</b><p>
<p style="line-height: 1.3em" class="details">
<div class="control-group">
<label class="control-label">ID </label>
<input type="text"name="studentID" align= "right" class="textbox" ><span id="errorID"></span> <br/>
</div>
<div class="control-group">
<label class="control-label" >First name</label>
<input type="text" name="Fname" class="textbox" ><span id="errorFirstName"> <br/>
</div>
</form>
How do I set the retrived data to be loaded in to the form's input fields?
What I have is a query to retrieve the ID of the record but I don't know how to set the entire query result on to the fields.
my php query:
<?php
if(isset($_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
php html mysql html5
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
add a comment |
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
add a comment |
2 Answers
2
active
oldest
votes
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' ] ) ? $row[ 'your_filed_in_database' ] : '' ?>" />
<input type="submit" name="send" />
</form>
Thanks for the example! However, I'm still stuck on understanding how to pass the query results to my form. shouldn't be a code for opening my desired page after the php code is done?
– user2674835
Apr 16 '14 at 16:00
add a comment |
You need to output layout of your form from php file.
Here is small example can help you.
File register.php
if (!isset($_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
}
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
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>
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
add a comment |
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>
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
add a comment |
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>
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' ] ) ? $row[ 'your_filed_in_database' ] : '' ?>" />
<input type="submit" name="send" />
</form>
answered Apr 16 '14 at 14:59
raiserleraiserle
424420
424420
Thanks for the example! However, I'm still stuck on understanding how to pass the query results to my form. shouldn't be a code for opening my desired page after the php code is done?
– user2674835
Apr 16 '14 at 16:00
add a comment |
Thanks for the example! However, I'm still stuck on understanding how to pass the query results to my form. shouldn't be a code for opening my desired page after the php code is done?
– user2674835
Apr 16 '14 at 16:00
Thanks for the example! However, I'm still stuck on understanding how to pass the query results to my form. shouldn't be a code for opening my desired page after the php code is done?
– user2674835
Apr 16 '14 at 16:00
Thanks for the example! However, I'm still stuck on understanding how to pass the query results to my form. shouldn't be a code for opening my desired page after the php code is done?
– user2674835
Apr 16 '14 at 16:00
add a comment |
You need to output layout of your form from php file.
Here is small example can help you.
File register.php
if (!isset($_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
}
add a comment |
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
}
add a comment |
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
}
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
}
answered Nov 12 '16 at 11:48
Andrey YerokhinAndrey Yerokhin
21016
21016
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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