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Variable in MySQL query


mcfmullen

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Hello,

 

I am trying to pass a php variable into a MySQL select query and am not having any success. Here's my code:

<?php
$columns = 5;

$result = mysql_query("SELECT * FROM Animals WHERE Animals.introDateAnimal = '2010-06-03'");
$url='http://www.cool.com/animalpic/';

$numofrows = mysql_num_rows($result);
echo "<table>";

for($i = 0; $i < $numofrows; $i++) {
    $row = mysql_fetch_array($result);
    if($i % $columns == 0) {
        echo "<tr>";
    }
    echo "<td><img src='".$url.$row['photoAnimal']."' height='100px' width='100px'>";
    echo "<br>";
    echo "<a href='http://www.cool.com/animals.php?animal={$row['nameAnimal']}'>{$row['nameAnimal']}</a></td>";
    if(($i % $columns) == ($columns - 1) || ($i + 1) == $numofrows) {
        echo "</tr>";
    }
}
echo "</table>";
?>

 

The code above gives me no results. No errors, but no results either. If I hard code the variable as such:

<?php
$columns = 5;
$date = 2010-06-03;

$result = mysql_query("SELECT * FROM Animals WHERE Animals.introDateAnimal = '$date'");
$url='http://www.cool.com/animalpic/';

$numofrows = mysql_num_rows($result);
echo "<table>";

for($i = 0; $i < $numofrows; $i++) {
    $row = mysql_fetch_array($result);
    if($i % $columns == 0) {
        echo "<tr>";
    }
    echo "<td><img src='".$url.$row['photoAnimal']."' height='100px' width='100px'>";
    echo "<br>";
    echo "<a href='http://www.cool.com/animals.php?animal={$row['nameAnimal']}'>{$row['nameAnimal']}</a></td>";
    if(($i % $columns) == ($columns - 1) || ($i + 1) == $numofrows) {
        echo "</tr>";
    }
}
echo "</table>";
?>

 

I get the output that I want. So the question is, how do I get the variable (first code) to work?

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$date = 2010-06-03; is a mathematical expression and results in the number 2001 (2010 minus 6 minus 3.)

 

Use the following to assign the string consisting of the characters '2010-06-03' that represents a date value -

$date = '2010-06-03';

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