Why does using an If statement change a variable? - php

Question

I am reading a text file and processing some records, a relevant sample of the text file is
#export_dategenre_idapplication_idis_primary
#primaryKey:genre_idapplication_id
#dbTypes:BIGINTINTEGERINTEGERBOOLEAN
#exportMode:FULL
127667880285760063715151750
127667880285760123715151751
I want to perform a specific action when application_id is already stored within my database AND is_primary = 1
I wrote this PHP to test my code:
$fp1 = fopen('genre_application','r');
if (!$fp) {echo 'ERROR: Unable to open file.'; exit;}
while (!feof($fp1)) {
$line = stream_get_line($fp1,128,$eoldelimiter); //use 2048 if very long lines
if ($line[0] === '#') continue; //Skip lines that start with #
$field = explode ($delimiter, $line);
list($export_date, $genre_id, $application_id, $is_primary ) = explode($delimiter, $line);
// does application_id exist?
$application_id = mysql_real_escape_string($application_id);
$query = "SELECT * FROM jos_mt_links WHERE link_id='$application_id';";
$res = mysql_query($query);
if (mysql_num_rows($res) > 0 ) {
echo $application_id . "application id has genre_id" . $genre_id . "with primary of " . $is_primary. "\n";
} else
{
// no, application_id doesn't exist
}
} //close reading of genre_application file
fclose($fp1);
which results in this output on screen and is exactly as I expected.
371515175application id has genre_id6006with primary of 0
371515175application id has genre_id6012with primary of 1
If I then add an IF statement as in the code below, it somehow changes the value of is_primary as shown by the screen display
$fp1 = fopen('genre_application','r');
if (!$fp) {echo 'ERROR: Unable to open file.'; exit;}
while (!feof($fp1)) {
$line = stream_get_line($fp1,128,$eoldelimiter); //use 2048 if very long lines
if ($line[0] === '#') continue; //Skip lines that start with #
$field = explode ($delimiter, $line);
list($export_date, $genre_id, $application_id, $is_primary ) = explode($delimiter, $line);
// does application_id exist?
$application_id = mysql_real_escape_string($application_id);
$query = "SELECT * FROM jos_mt_links WHERE link_id='$application_id';";
$res = mysql_query($query);
if (mysql_num_rows($res) > 0 ) {
if ($is_primary = '1') echo $application_id . "application id has genre_id" . $genre_id . "with primary of " . $is_primary. "\n";
} else
{
// no, application_id doesn't exist
}
} //close reading of genre_application file
fclose($fp1);
?>
The code above results in the following screen display, which incorrectly has the first field with a primary of 1, when as can be seen by the previous screen display and the sample text file it should be 0
371515175application id has genre_id6006with primary of 1
371515175application id has genre_id6012with primary of 1
Can anyone explain what I am doing to make the variable change and how I should use the If correctly please?

Answer 1

You are assigning a value instead of comparing:
($is_primary = '1')
you need
($is_primary == '1')
or === for a type-safe comparison.
This is why some people like to write their comparisons like so:
('1' == $is_primary)
the mistake is impossible to make here because "1" can't be assigned anything.
Personally though, I think that over time and with growing practice, one will learn to spot the mistake.

Answer 2

You need to use:
if ($is_primary == '1')
NOT
if ($is_primary = '1')
Because "=" defines variable and returns true, but "==" actually compares and doesn't change anything.

Answer 3

The if-line has only one = (assign value) instead of two == (compare value).
Try this instead:
if ($is_primary == '1')

Answer 4

if ($is_primary = '1')
You need to use the comparison operator ==, not the assignment operator =

Answer 5

if ($is_primary == '1')
....
NOT = BUT ==
Also consider removing all the stuff that bug the reading of your question for example somthing like:
// does application_id exist?
$res = mysql_query($query);
if (mysql_num_rows($res) > 0 ) {
echo $is_primary. "\n";
}
This is more readable and fit you purpose.
You may even find yourself does kind of bug.

Answer 6

If you type
if($x=10)...
Then you're telling it to attempt the operation
$x=10
to which it will return a success (return true). This will always run the loop, and alter the value of x.
To test $x, use
if($x==10)...
But note that to test for $x not equals 10 is
if($x!=10)...
ie, one equals and one bang.

Answer 7

Chek out PHP comparison operators