i want to try fetch data on other page and than update but always show me an error

Question

here is my index page.inserted all the data to the database and also show on the same page but the main problem is that on update.php page I can not retrieve the data

//that main problem is here and I can't be retrieved the data on this page and always sow that: Warning: mysql_fetch_array() expects parameter 1 to be resource, object given in C:\wamp\www\phonebook\update.php on line 12

                index.php


                <?php require_once('dbconnect.php'); ?>

                <html>
                <head>
                <title> </title>
                </head>
                <body>
                <h1> phone book </h1>

                <form method="post">
                <table>
                <tr> 
                <td>fname </td><td> <input type="text" name="firstname" required  /> </td>
                </tr>
                <tr> 
                <td>lname </td><td> <input type="text" name="lastname" required  /> </td>
                </tr>
                <tr> 
                <td>mobile </td><td> <input type="text" name="mobile" required  /> </td>
                </tr>
                </table>    
                <input type="submit" name="submit" value="submit" >

                </form> 

                <!-- $$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$ show  $$$$$$$$$$$$$$$$$$$$$$$$$$ -->

                <br> data </br>
                <table border="1">
                <tr>
                <th>id</th>   <th>firstname</th>      <th>lastname</th> <th>mobile</th><th>update</th><th>delete</th>
                </tr>
                <?php
                $conn = mysqli_connect('localhost','root','','phonebook');
                $show = mysqli_query($conn,"SELECT * FROM contacts");

                while($row = mysqli_fetch_array($show)) 
                {
                ?>

                <tr> 
                <td><?php echo $row['id']; ?></td>
                <td><?php echo $row['firstname']; ?></td>
                <td><?php echo $row['lastname']; ?></td>
                <td><?php echo $row['mobile']; ?></td>
                <td><a href="update.php?id=<?php echo $row['id']; ?>">update</a></td>
                <td><a href="delete.php?id=<?php echo $row['id']; ?>" onclick="return confirm('sure want to delete')" >delete</a></td>
                </tr>


                <?php }  ?>
                </table>
                </body>
                </html>

                <?php
                //require_once("function.php");
                //$obj = new data();


                if(isset($_POST{"submit"}))
                {

                //echo "<pre>";print_r($_POST);die;
                $fname = $_POST['firstname'];
                $lname = $_POST['lastname'];
                $mobile = $_POST['mobile'];
                //$obj->insert($fname,$lname,$mobile);
                $connect = mysqli_connect('localhost','root','','phonebook');
                $insert = mysqli_query($connect,"insert into contacts(firstname,lastname,mobile) values('".$fname."','".$lname."','".$mobile."')");

                if ($insert)
                {  ?>

                <script> alert('record inserted'); </script>
                <?php 
                }
                else
                { ?>
                <script> alert('record not inserted'); </script>
                <?php

                }

                header('Location:index.php');
                }

                ?>

                update.php  
//check the code here 

                <?php require_once('dbconnect.php'); 

                if(isset($_GET['id']) && is_numeric($_GET['id']) ) 
                {
                $id=$_GET['id'];
                }

                ?>
                <?php
                $conn = mysqli_connect('localhost','root','','phonebook');
                $result=mysqli_query($conn,"SELECT * FROM contacts WHERE id='$id'");
                $fetch=mysql_fetch_array($result);
                //$conn = mysqli_connect('localhost','root','','phonebook');
                //$show = mysqli_query($conn,"SELECT * FROM contacts");

                //while($row = mysqli_fetch_array($show)) 


                ?>


                <html>
                <head>
                <title>update page</title>
                </head>
                <body>
                <form method="post"  name="update" action="update.php">
                <table>

                <tr> 
                <td>fname </td><td> <input type="text" name="firstname" value= "<?php echo $fetch['firstname']; ?>" required  /> </td>
                </tr>
                <tr> 
                <td>lname </td><td> <input type="text" name="lastname" value="<?php echo $fetch['lastname']; ?>" required  /> </td>
                </tr>
                <tr> 
                <td>mobile </td><td> <input type="text" name="mobile" value= "<?php echo $fetch['mobile']; ?>" required  /> </td>
                </tr>
                </table>    
                <input type="submit" name="submit" value="submit" >

                </form> 

                </body>
                </html>

Possible duplicate of [Can I mix MySQL APIs in PHP?](http://stackoverflow.com/questions/17498216/can-i-mix-mysql-apis-in-php) — u_mulder, Mar 17 '17 at 11:42
I have seen a similar problem. The answer is in here : http://stackoverflow.com/questions/2973202/mysqli-fetch-array-mysqli-fetch-assoc-mysqli-fetch-row-expects-parameter-1 — Marouane Sihad, Mar 17 '17 at 11:48

score 1 · Accepted Answer · answered Mar 17 '17 at 11:43

1

Switch to using mysqli_fetch_array() (note the i) instead of mysql_fetch_array

answered Mar 17 '17 at 11:43

Scuzzy

12,186
1
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score 1 · Answer 2 · answered Mar 17 '17 at 11:49

try this:

$conn = mysqli_connect('localhost','root','','phonebook');
$result=mysqli_query($conn,"SELECT * FROM contacts WHERE id='$id'");
$fetch=mysqli_fetch_array($result);

You must not use mysql_*, it's deprecated. Use PDO or MySQLi instead
You shouldn't mix mysql_* and mysqli_*
Just create ONE mysqli instance instead of creating it for every file you have.
Maximize the use of variables too. This way you only have to change something once.
Please sanitize/escape user input before passing it into your SQL query. Otherwise your application is vulnerable to SQL injection attacks.

i want to try fetch data on other page and than update but always show me an error

2 Answers2