If variable equals value php
67,984
Solution 1
You are comparing, not assigning:
if ($type == 1){
$type = "Bear";
}
You compare values with ==
or ===
.
You assign values with =
.
You could write less code to achieve the same result too, with a switch
statement, or just a bunch of if
s without the elseif
s.
if ($type == 1) $type = "Bear";
if ($type == 2) $type = "Cat";
if ($type == 3) $type = "Dog";
I would make a function for it, like this:
function get_species($type) {
switch ($type):
case 1: return 'Bear';
case 2: return 'Cat';
case 3: return 'Dog';
default: return 'Jeff Atwood';
endswitch;
}
$type = get_species($row['ttype']);
Solution 2
You are using ==
instead of =
. It compares the variable to the new value. Use =
to set the value.
if ($type == 1){
$type = "Bear";
} elseif ($type == 2){
$type = "Cat";
} elseif ($type == 3){
$type = "Dog";
}
Solution 3
You're using ==
to assign values:
$type == bear;
Should be:
$type = bear;
Author by
llw
Updated on July 12, 2022Comments
-
llw almost 2 years
I am trying to do a check before the data inserts into the MySQL query. Here is the code;
$userid = ($vbulletin->userinfo['userid']); $sql3 = mysql_query("SELECT * FROM table WHERE ID='$_POST[hiddenID]'"); while ($row = mysql_fetch_array($sql3)){ $toon = $row['toonname']; $laff = $row['tlaff']; $type = $row['ttype']; if ($type == 1){ $type == "Bear"; } elseif ($type == 2){ $type == "Cat"; } elseif ($type == 3){ $type == "Dog"; } }
However, this isn't working. Basically, there are different values in the 'table' for each type. 1 means Bear, 2 means Cat, and 3 means Dog.
Thanks to whomever can help see a problem in my script!