I've got a php script with this statement:
开发者_运维技巧$query = "SELECT type
FROM users
WHERE username = '$username'";
$result = $database->query($query);
if($result == 1) {
echo "whatever";
continue;
}
The problem is that the if never runs and when I created a print statement to print $result before the if runs, it prints a Reference ID#. So result is never == 1 because it is being assigned the reference ID #.
What the heck am I doing wrong? How do I assign the value of 'type' which is an INT, instead of it's contents Reference ID#?
you have to fetch that line first ...
$query = "SELECT type FROM users WHERE username = '$username'";
$result = $database->query($query);
$row = mysql_fetch_assoc($result);
if($row['type'] == 1)
{
echo "whatever";
}
You need to use mysql_fetch_assoc() or mysql_fetch_array() to get the result into an array:
$query = $database->query("SELECT type FROM users WHERE username = '$username'");
$result = mysql_fetch_assoc($query);
if($result['type'] == 1)
{
echo "whatever";
continue;
}
try
<?php
$result = mysql_fetch_assoc($database->query($query));
$id = $result['type'];
if($type == 1)
{
Yes. You have to check it like:
$query = "SELECT type FROM users WHERE username = '$username'";
$result = $database->query($query);
if($result)
{
echo "whatever";
continue;
}
This is a Boolean check, but... In php we have weak types - everything is true, except on false, 0, '' (empty string), null, etc.
If you want to make a good Boolean check including type, then you have:
if(bool_val === true)
{
..
}
// or
if(bool_val && is_bool(bool_val))
{
..
}
This query object returns a result resource in php. And you probably receive #Resource ID, which is some kind of pointer in php code.
If you receive something like this (#Resource ID) - this means your query has passed correctly and no error occured.
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