sql data fetch bug

Question

I have two pages . In the first page if user will click on button then he will get relavant data with id column

<form action='edit.php?id=$id' method='post' name='edit_btn'> 
<button type='submit' class='w3-btn w3-red w3-round-xlarge'>Proceed </button> 
</form>   

        $con=mysqli_connect("localhost","root","","student_result_info_db");if (mysqli_connect_errno()){ echo "Failed to connect to MySQL: " . mysqli_connect_error(); }else{ 
          if(isset($_GET['edit_btn'])){
              echo "<div class='w3-container w3-red'> <h1>>Error Found!</h1> <h4>OK You Did Hit The EDIT Button</h4> </div> <br/>";
          }else{  
             $get_selected_id = $_GET['id'];
             echo $get_selected_id;

             $res = "SELECT * FROM school_result";


                 while ($row = mysql_fetch_array($res)){
                     echo "$row[id]. $row[first_name] <br/>";
                     echo "$row[id]. $row[last_name] <br/>";
                 } 
          }
    }

It'showing following error

Fatal error: Call to undefined function mysql_fetch_array() in

Are you using PHP 7? `mysql_fetch_array` was removed in PHP 7 and deprecated in PHP 5. — Aaron Christiansen, Mar 17 '16 at 12:45
You are using `mysqli` to connect but try to use `mysql` to fetch the data, you can't mix match like that. — Epodax, Mar 17 '16 at 12:46
@Epodax I using php 7 . Please give me correct code for this bug — Gourab Mazumder, Mar 17 '16 at 12:48
@GourabMazumder It's all ready been provided, but just a FYI, it's **not** a bug, your code is wrong. - And also, you need to execute your query, do please find a tutorial on php / mysql and learn it, your code is showing a lack of understanding. — Epodax, Mar 17 '16 at 12:53
@OrangeFlash81 Now I using `while ($row = mysqli_fetch_array(mysqli_query($con, $res))){ ` It's working but it's fetchng all data from database but i I need fetch which id is maching — Gourab Mazumder, Mar 17 '16 at 13:07

Aaron Christiansen · Accepted Answer · 2016-03-17T13:03:35.207

0

Use mysqli_fetch_array instead, along with mysqli_query to get a mysqli_result object:

$qresult = mysqli_query($con, $res);
while ($row = mysqli_fetch_array($qresult)){

You need to use this for 2 reasons:

You opened your database using mysqli_connect, so you need to use the mysqli functions.
mysql functions were removed in PHP 7.

edited Mar 17 '16 at 13:03

answered Mar 17 '16 at 12:47

Aaron Christiansen

11,584
5
52
78

it's hsowing error **Warning: mysqli_fetch_array() expects parameter 1 to be mysqli_result, string given in** – Gourab Mazumder Mar 17 '16 at 12:52
@GourabMazumder Sorry: I'll change my answer. – Aaron Christiansen Mar 17 '16 at 12:53
@GourabMazumder I've changed my answer to use `mysqli_query`. – Aaron Christiansen Mar 17 '16 at 12:55
It's fetching but here is a problem . In the db many rows of data was inserted and now **I need fetch only which row that column `ïd` is matching** and it also showing a error **Fatal error: Maximum execution time of 30 seconds exceeded in** – Gourab Mazumder Mar 17 '16 at 13:27

score 0 · Answer 2 · answered Mar 17 '16 at 12:47

0

Use mysqli function instead of mysql like

mysqli_query($con, $res);

answered Mar 17 '16 at 12:47

Jatin Raikwar

406
5
17

it's show error **Warning: mysqli_fetch_array() expects parameter 1 to be mysqli_result, string given in** – Gourab Mazumder Mar 17 '16 at 12:53
have you used mysqli_query function – Jatin Raikwar Mar 17 '16 at 12:56
It's fetching but here is a problem . In the db many rows of data was inserted and now **I need fetch only which row that column `ïd` is matching** and it also showing a error **Fatal error: Maximum execution time of 30 seconds exceeded in** – Gourab Mazumder Mar 17 '16 at 13:28

sql data fetch bug

2 Answers2