CODEWITHSUNDEEP
phpmysql
Lukas Pleva
Lukas Pleva

Reputation: 381

Editing data from MySQL via PHP

I am running into a frustrating problem with a PHP script that's supposed to allow me to edit individual rows within my MySQL database.

This is the file where all of the rows from the database are displayed; it works just like it's supposed to. 

<table cellpadding="10">
<tr>
<td>ID</td>
<td>First Name</td>
<td>Last Name</td>
<td>E-mail</td>
<td>Phone</td>
</tr>

<?php

$username="username here";
$password="password here";
$database="database name here";

mysql_connect(localhost,$username,$password);
@mysql_select_db($database) or die( "Unable to select database");
$query="SELECT * FROM students";
$result=mysql_query($query);
mysql_close();

while ($row=mysql_fetch_array($result)){
echo ("<tr><td>$row[id]</td>");
echo ("<td>$row[first]</td>");
echo ("<td>$row[last]</td>");
echo ("<td>$row[email]</td>");
echo ("<td>$row[phone]</td>");
echo ("<td><a href=\"StudentEdit.php?id=$row[id]\">Edit</a></td></tr>");
}
echo "</table>";

?>

As you can see, each row has an "Edit" link that is supposed to allow the user to edit that individual student's data. Here, then, is StudentEdit.php:

<?php

$username="username";
$password="password";
$database="database";

mysql_connect(localhost,$username,$password);

$student_id = $_GET[id]; 
$query = "SELECT * FROM students WHERE id = '$student_id'"; 
$result = mysql_query($query);
$row = mysql_fetch_array($result);
mysql_close();
?>

<form method="post" action="EditStudentData.php" />

<table>

<tr>
<td><input type="hidden" name="id" value="<? echo "$row[id]" ?>"></td>
</tr>

<tr>
<td>First Name:</td>
<td><input type="text" name="first" value="<? echo "$row[first]" ?>"></td>
</tr>

<tr>
<td>Last Name:</td>
<td><input type="text" name="last" value="<? echo "$row[last]" ?>"></td>
</tr>

<tr>
<td>Phone Number:</td>
<td><input type="text" name="phone" value="<? echo "$row[phone]" ?>"></td>
</tr>

<tr>
<td>E-mail:</td>
<td><input type="text" name="email" value="<?echo "$row[email]" ?>"></td>
</tr>

</table>

</form>

When I execute this, however, I get the following error message:

Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in /home4/lukaspl1/public_html/StudentEdit.php on line 12

Any ideas what's wrong, and how to fix it?

Thank you in advance!

Upvotes: 4

Views: 50295

Answers (9)

Dasun Hansaka
Dasun Hansaka

Reputation: 81

This part of the code is wrong:

$student_id = $_GET[id];

the correct code is

$student_id = $_GET['id'];

code from expertsnote.com

Upvotes: 2

indranatha madugalle
indranatha madugalle

Reputation: 13

This code gives the option to add, search, edit and delete options. Thought it might to see all the options in one code. 

        $searchedUsername = "";
        $searchedEmail = "";
    //registration (Add) function   
        if ( isset($_POST['stdregister'])){

            $username = $_POST['stdusername'];
            $password = $_POST['stdpassword'];
            $email = $_POST['stdemail'];
            $hashedPassword = md5($password);

            $connection = mysqli_connect("localhost","root","","std");
            $query = "INSERT INTO student VALUES ('$username','$hashedPassword','$email')";
            if ( mysqli_query($connection,$query) == 1 ){
                echo "Successfully saved";
            }
            else{
                echo "<p style='color: #f00;'>There is an error</p>";
            }
            mysqli_close($connection);
        }

    //delete function   
        if ( isset($_POST['stddelete'])){
            $username = $_POST['stddelusername'];
            $connection = mysqli_connect("localhost","root","","std");
            $query = "DELETE FROM student WHERE username LIKE '$username'";
            mysqli_query($connection,$query);
            echo mysqli_error($connection);

            mysqli_close($connection);
        }
    //update function   
        if ( isset($_POST['stdupdate'])){
            $username = $_POST['stdusername'];
            $stdpass = md5($_POST['stdpassword']);
            $stdemail = $_POST['stdemail'];
            $connection = mysqli_connect("localhost","root","","std");
            $query = "UPDATE student SET password='$stdpass', email='$stdemail' WHERE username LIKE '$username'";
            mysqli_query($connection,$query);
            echo mysqli_error($connection);
            mysqli_close($connection);
        }

        if ( isset($_POST['stdsearch']) ){
            $searchUsername = $_POST['stdeditusername'];
            $connection = mysqli_connect("localhost","root","","std");
            $query = "SELECT * FROM student WHERE username LIKE '$searchUsername' ";
            $result = mysqli_query($connection, $query);
            while( $row = mysqli_fetch_array($result) ){
                $searchedUsername = $row['username'];
                $searchedEmail = $row['email'];
            }
        }

    ?>
    <html>
        <head>


        </head>
        <body>
            <h1>Student Registration</h1>
            <form name="stdregistration" action="forms.php" method="post">
                <label>Username :</label>
                <input name="stdusername" required="required" type="text" /><br /><br />
                <label>Password :</label>
                <input name="stdpassword" type="password" /><br /><br />
                <label>E-mail :</label>
                <input name="stdemail" type="email" /><br /><br />
                <input name="stdregister" type="submit" value="Save" />
            </form>

            <h2>Delete Students</h2>
            <form name="stddeletion" action="forms.php" method="post">
                <label>Select the Username :</label>
                    <select name="stddelusername" required>
                    <option value="">Select One</option>
    <?php
        $connection2 = mysqli_connect("localhost","root","","std");
        $query2 = "SELECT username FROM student";
        $result = mysqli_query($connection2,$query2);
        while( $row = mysqli_fetch_array($result) ){
            echo "<option value='".$row['username']."'>".$row['username']."</option>";
        }
        mysqli_close($connection2);

    ?>          


                </select>
                <input name="stddelete" type="submit" value="Delete" />
            </form>

            <h2>Edit Students</h2>
            <form name="stdedition" action="forms.php" method="post">
                <label>Select the Username :</label>
                    <select name="stdeditusername" required>
                    <option value="">Select One</option>
    <?php
        $connection2 = mysqli_connect("localhost","root","","std");
        $query2 = "SELECT username FROM student";
        $result = mysqli_query($connection2,$query2);
        while( $row = mysqli_fetch_array($result) ){
            echo "<option value='".$row['username']."'>".$row['username']."</option>";
        }
        mysqli_close($connection2);

    ?>          
                </select>
                <input name="stdsearch" type="submit" value="Search" />
            </form>

            <form name="stdedit" action="forms.php" method="post">
                <label>Username :</label>
                <input name="stdusername" required="required" type="text" readonly value="<?php echo $searchedUsername; ?>" /><br /><br />
                <label>Password :</label>
                <input name="stdpassword" type="password" /><br /><br />
                <label>E-mail :</label>
                <input name="stdemail" type="email" value="<?php echo $searchedEmail; ?>" /><br /><br />
                <input name="stdupdate" type="submit" value="Update" />
            </form>
        </body>
    </html>

Upvotes: 0

user9232865
user9232865

Reputation:

this is the cod for edit the details dynamically 

<?php

    include('db.php');

        $id=$_REQUEST['id'];
         $query="SELECT * FROM `camera details` WHERE id='".$id."'";
        $result=mysqli_query($db,$query) or die(mysqli_error());
        $row1=mysqli_fetch_assoc($result);


    if(isset($_POST['submit'])&&(isset($_POST['new'])&&($_POST['new'])==1))
    {
            $id=$_REQUEST['id'];

        foreach($_POST as $key=>$values)
        {
            if($key!="submit"){

            $names[]=$key;
            $val[]= "'".$values."'";

            if($key!="new"){
            $k[] = "`".$key."` = '".$values."'";
            }
            }
        }
        $output=implode(",",(array)($k));
        //$v=implode(",",(array)($val));

        // `name` = 'san'

        $query="UPDATE `camera details` SET $output WHERE id='".$id."'";

        $output=mysqli_query($db,$query) or die(mysqli_error($db));
        if($output)
        {
            header('location:cameralist.php');
        }
    }
    else{
        ?>

Upvotes: 1

Dasun Hansaka
Dasun Hansaka

Reputation: 81

this code was missing

$select_db = mysql_select_db("$db_name");
if (!$select_db) {echo "Error Selecting Database";}

Upvotes: 1

Ashwini Agarwal
Ashwini Agarwal

Reputation: 4858

Try...

echo ("<td><a href=\"StudentEdit.php?id=".$row['id']."\">Edit</a></td></tr>");

instead of 

echo ("<td><a href=\"StudentEdit.php?id=$row[id]\">Edit</a></td></tr>");

Upvotes: 1

ryuusenshi
ryuusenshi

Reputation: 1986

StudentEdit.php: you forgot to call @mysql_select_db($database) or die( "Unable to select database"); before you executed the query

Upvotes: 2

Conrad Lotz
Conrad Lotz

Reputation: 8838

What looks incorrect to me is the way you are displaying the value retrieved from the database:

<input type="hidden" name="id" value="<? echo "$row[id]" ?>">

It should be

<input type="hidden" name="id" value="<?php echo $row['id']; ?>">

Upvotes: 0

Dale
Dale

Reputation: 10469

I recommend doing this in studentEdit.php

$student_id = mysql_real_escape_string($_GET[id]); 
$query = "SELECT * FROM students WHERE id = '$student_id'"; 
$result = mysql_query($query) or die(mysql_error() . ' ' . $query);
$row = mysql_fetch_array($result);
mysql_close();

Two things I've changed here is firstly to escape the data being passed in the url and secondly I've added or die(mysql_error() . ' ' . $query); If something is going wrong in the sql statement you should now see the error and hopefully you'll be able to fix it from there.

Upvotes: 0

satdev86
satdev86

Reputation: 810

Remove the mysql_close from here

mysql_connect(localhost,$username,$password);
@mysql_select_db($database) or die( "Unable to select database");
$query="SELECT * FROM students";
$result=mysql_query($query);
mysql_close();

The code should mysql_connect(localhost,$username,$password);

@mysql_select_db($database) or die( "Unable to select database");
$query="SELECT * FROM students";
$result=mysql_query($query);

And moreover,you are going to use only key based resultset.. simply have mysql_fetch_assoc. And another suggestion would be instead of $row[id]..replace it with $row['id'].

Upvotes: 3

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