在php和mysql中链接[重复]

Question

This question already has an answer here:

• mysql_fetch_array()/mysql_fetch_assoc()/mysql_fetch_row()/mysql_num_rows etc… expects parameter 1 to be resource or result 32 answers

I load data from mysql database and display it into listview jqm. Here is my code:

<?php
       include "connect.php";
        $result = mysqli_query($con,"SELECT * FROM story");
        id==$row['id'];
        echo "<ul data-role='listview' data-inline='true'>";
        while($row = mysqli_fetch_array($result)) {
          echo "<li>" ;
          echo "<a href='view.php?id=".$rows['id']."' data-rel='external'>";
          echo $row['name'];
          echo "</a>";
          echo "</li>";
            }
        echo "</ul>";
        ?>

and view page :

<?php
    include "connect.php";
    $id = $_GET['id'];
    $result = mysqli_query($con,"SELECT * FROM story WHERE id='".$id."'");
    $row = mysqli_fetch_array($result);
    echo $row['content'];

    ?>

But it not working and this error: Warning: mysqli_fetch_array() expects parameter 1 to be mysqli_result, boolean given in C:\xampp\htdocs\1\view.php on line 11. What should I do ?

</div>

Answer 1

This error shows that the SQL Query is unable to get data into $result and if you try to print this result on the screen it will show you bool(false) as output. This parameter when passed into the mysqli_fetch_array() does not fetch array for you because it requires a TRUE boolean value.

First try to get the data from the database into the $result, print it and then proceed forward.

Maybe this will help,

$result = mysqli_query($con,"SELECT * FROM story WHERE id='.$id.' ");

Remember to always use error checking methods like die, mysqli_error, mysqli_connect_error(), mysqli_connect_errno() etc for better debugging.