Cant output correct result to variable

Question

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', '&lt;?php', $snip);
$snip = str_replace('?>', '?&gt;', $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);

score 0 · Accepted Answer · answered Sep 06 '14 at 00:39

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', '&lt;?php', $snip);
$snip = str_replace('?>', '?&gt;', $snip);
?>
    <div id="editor"><?php echo $snip; ?></div>

Cant output correct result to variable

1 Answers1