I have problem in retrieving record from mysql using php form, I was able to retrieve the record, if I edit, the changes is not updated in mysql.
find below my code -
<html>
-
<body>
-
<?php
-
$db = mysql_connect("localhost", "root");
-
mysql_select_db("k1",$db);
-
$id = $_POST["id"];
-
echo $id;
-
if ($submit) {
-
$first = $_POST['first'];
-
$last = $_POST['last'];
-
$address = $_POST['address'];
-
$position = $_POST['position'];
-
$sql = "UPDATE 'employees' SET 'first'='$first','last'='$last','address'='$address','position'='$position' WHERE 'id'='$id'";
-
$result = mysql_query($sql);
-
echo $first;
-
echo "Thank you! Information updated.\n";
-
} else {
-
if ($id) {
-
$sql = "SELECT * FROM employees WHERE id=$id";
-
$result = mysql_query($sql);
-
$myrow = mysql_fetch_array($result);
-
$first = $myrow['1'];
-
$last = $myrow['2'];
-
$address = $myrow['3'];
-
$position = $myrow['4'];
-
$id = $myrow['id'];
-
$submit = 'submit';
-
}
-
}
-
?>
-
<form method="post" action="<?php echo $PHP_SELF?>">
-
<input type=hidden name="id" value="<?php echo $myrow["id"] ?>">
-
First name:<input type="Text" name="first" value="<?php echo $first ?>"><br>
-
Last name:<input type="Text" name="last" value="<?php echo $last ?>"><br>
-
Address:<input type="Text" name="address" value="<?php echo $address ?>"><br>
-
Position:<input type="Text" name="position" value="<?php echo $position ?>"><br>
-
<input type="Submit" name="submit" value="Enter information">
-
</form>
-
</body>
-
</html>
[Please use CODE tags when posting source code. Thanks! --pbmods] 4  2219  Purple 404
Recognized Expert Contributor
Hi Bakarre and welcome to TSDN !!
Can you take a look at this excellent post by Motoma where he works through options to get debug output from PHP. I suspect from looking at your code you have not got error messages turned on.
Purple
 Purple 404
Recognized Expert Contributor
And try this
[PHP]<html>
<body>
<?php
$db = mysql_connect("localhost", "root");
mysql_select_db("k1",$db);
if(isset($_POST["id"])) $id = $_POST["id"];
//echo $id;
if (isset($_POST['submit']))
{
$first = $_POST['first'];
$last = $_POST['last'];
$address = $_POST['address'];
$position = $_POST['position'];
$sql = "UPDATE 'employees' SET 'first'='$first','last'='$last','address'='$addres s','position'='$position' WHERE 'id'='$id'";
$result = mysql_query($sql);
echo $first;
echo "Thank you! Information updated.\n";
}
elseif (isset($id))
{
$sql = "SELECT * FROM employees WHERE id=$id";
$result = mysql_query($sql);
$myrow = mysql_fetch_array($result);
$first = $myrow['1'];
$last = $myrow['2'];
$address = $myrow['3'];
$position = $myrow['4'];
$id = $myrow['id'];
$submit = 'submit';
}
?>
<form method="post" action="<?php echo $PHP_SELF?>">
<input type=hidden name="id" value="<?php echo $myrow["id"] ?>">
First name:<input type="Text" name="first" value="<?php echo $first ?>"><br>
Last name:<input type="Text" name="last" value="<?php echo $last ?>"><br>
Address:<input type="Text" name="address" value="<?php echo $address ?>"><br>
Position:<input type="Text" name="position" value="<?php echo $position ?>"><br>
<input type="Submit" name="submit" value="Enter information">
</form>
</body>
</html>[/PHP]
Let me know how you get on
Purple
Hi,
is perfectly now working.
You can buy experience.
I gained alot.
Bakare
Hi,
is perfectly now working.
You cann't buy experience.
I gained alot.
Bakarre
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