基本的PHP SQL查询无法正常工作

Question

We have a basic PHP script to extract the title and description for each job from a MySQL database as simply display this information. This is what it looks like:

$sql        = "SELECT `title`, `desc` FROM jobs WHERE active = 'y'";
$query      = mysql_query($sql) or die('<em><strong>SQL Error:</strong> ' . mysql_error() . '</em>');
$results    = mysql_fetch_assoc($query);

<?php while($result = mysql_fetch_assoc($query)) {
    echo '<div class="left_content" style="margin-top: 15px;">';
    echo "<h2>{$results['title']}</h2>";
    echo "<p>{$results['desc']}</p>";
    echo '</div>';
} ?>

Now, this only extracts one row from the database, but it should extract two. So, I tried the following to replace the while statement:

<?php foreach($results as $result) {
    echo '<div class="left_content" style="margin-top: 15px;">';
    echo "<h2>{$result['title']}</h2>";
    echo "<p>{$result['desc']}</p>";
    echo '</div>';
} ?>

This statement doesn't work either. This just displays (weirdly) the first character of each column in the first row in the table.

Does anyone have any idea as to why this isn't working as it should?

Answer 1

In your while use same variable $result as you started:

while($result = mysql_fetch_assoc($query)) {
    echo '<div class="left_content" style="margin-top: 15px;">';
    echo "<h2>{$result['title']}</h2>";
    echo "<p>{$result['desc']}</p>";
    echo '</div>';
}

and remove the first $results = mysql_fetch_assoc($query);

Answer 2

Result variable you have used is result not results

Replace

$sql        = "SELECT `title`, `desc` FROM jobs WHERE active = 'y'";
$query      = mysql_query($sql) or die('<em><strong>SQL Error:</strong> ' . mysql_error() . '</em>');
**$results    = mysql_fetch_assoc($query);** // remove this line

<?php while($result = mysql_fetch_assoc($query)) {
    echo '<div class="left_content" style="margin-top: 15px;">';
    echo "<h2>{$results['title']}</h2>";
    echo "<p>{$results['desc']}</p>";
    echo '</div>';
} ?>

to

$sql        = "SELECT `title`, `desc` FROM jobs WHERE active = 'y'";
$query      = mysql_query($sql) or die('<em><strong>SQL Error:</strong> ' . mysql_error() . '</em>');


<?php while($result = mysql_fetch_assoc($query)) {
    echo '<div class="left_content" style="margin-top: 15px;">';
    echo "<h2>{$result['title']}</h2>";
    echo "<p>{$result['desc']}</p>";
    echo '</div>';
} ?>

Answer 3

You already fetched the first row before your loop started, which is why it only prints the second row. Simply comment out that line:

#$results = mysql_fetch_assoc($query); # here is your first row, 
                                       # simply comment this line

<?php while($result = mysql_fetch_assoc($query)) {
    echo '<div class="left_content" style="margin-top: 15px;">';
    echo "<h2>{$result['title']}</h2>";
    echo "<p>{$result['desc']}</p>";
    echo '</div>';
} ?>

You are also looping over $result but using $results in your while loop body.

Answer 4

Check this:

$sql        = "SELECT `title`, `desc` FROM jobs WHERE active = 'y'";
$query      = mysql_query($sql) or die('<em><strong>SQL Error:</strong> ' . mysql_error() . '</em>');


<?php 
while($result = mysql_fetch_assoc($query)) {
    echo '<div class="left_content" style="margin-top: 15px;">';
    echo "<h2>{$result['title']}</h2>";
    echo "<p>{$result['desc']}</p>";
    echo '</div>';
}
 ?>

Answer 5

Change this line

<?php while($result = mysql_fetch_assoc($query)) {

to

<?php while($results = mysql_fetch_assoc($query)) {