无法将正确的结果输出到变量[重复]
问题描述：
This question already has an answer here:
right now im trying to append a result to a variable in php from a mysql query, this is the code used:
<?php
$con=mysqli_connect("localhost","root","","ezsnip");
// Check connection
if (mysqli_connect_errno()) {
echo "Failed to connect to MySQL: " . mysqli_connect_error();
}
$result = mysql_query("SELECT Snip FROM snips WHERE ID='1'");
$row = mysql_fetch_array($result);
$snip = $row['Snip'];
$snip = str_replace('<?php', '<?php', $snip);
$snip = str_replace('?>', '?>', $snip);
?>
<div id="editor"><?php echo $snip; ?></div>
When i run the code it returns: mysql_fetch_array() expects parameter 1 to be resource, boolean given in C:\xampp\htdocs\home.php on line 170
Line 170 is:
$row = mysql_fetch_array($result);
</div>
答
You're doing mysql_query after creating a mysqli resource
<?php
$con=mysqli_connect("localhost","root","","ezsnip");
// Check connection
if (mysqli_connect_errno()) {
echo "Failed to connect to MySQL: " . mysqli_connect_error();
}
$result = mysqli_query($con,"SELECT Snip FROM snips WHERE ID='1'");
$row = mysqli_fetch_array($result);
$snip = $row['Snip'];
$snip = str_replace('<?php', '<?php', $snip);
$snip = str_replace('?>', '?>', $snip);
?>
<div id="editor"><?php echo $snip; ?></div>