在sqlite数据库的不同页面上显示具有相似ID的数据

Question

Am new to php and i have successfully displayed data from sqlite database. The $catname is suppose to link to a page that show data from another table in the databse with the same id . Below is the code:

 <?php
require_once ("db.php");
$db = new MyDb();

    $sql =<<<EOF
SELECT * FROM addcategory ORDER BY catID DESC;
EOF;

    $ret = $db->query($sql);

    while ($row = $ret->fetchArray(SQLITE3_ASSOC)) {
        $catname = $row['catname'];
        $catdes = $row['catbrief'];
        $catimage = $row['catpic'];
        $catid = $row['catID'];

        echo "<div class=\"catDescription\">
<div class=\"catname\"><p><a href='search.php?category_id=$catid'>$catname</a>  </p></div>
    <div class=\"catImage\"><img src='".$catimage."'></div>
    <div class=\"catprof\"><p>$catdes</p></div>
    </div>";
    }

?> 

How do i display the data with thesame id with id from the other table

I have tried this on the other page

 <?php

$sql =<<<EOF
SELECT * FROM questions ORDER BY category_id DESC;
EOF;

$ret = $db->query($sql);

while ($row = $ret->fetchArray(SQLITE3_ASSOC)) {
    $catquestion = $row['question'];
    $catans = $row['answer'];


        echo "<div class=\"catDescription\">
<div class=\"catname\"><p>$catquestion</p></div>
    <div class=\"catprof\"><p>$catans</p></div>
    </div>";
}

?>

But it displays all data in the table and not data with similar id. Please what's wrong here and how do i fix this.

Answer 1

It seems you are missing the appropriate WHERE clause. Your query should be something like:

$cat_id = (int)$_GET['category_id']; //cast to int to strip out sql injections
$sql =<<<EOF
SELECT * FROM questions WHERE category_id = {$cat_id} ORDER BY category_id DESC;
EOF;