Checking if value exists in php array - not working?

Question

I have the following php code that gets a serialized array from a mysql database, then unserializes it. This is working fine. The following code:

$row=mysql_fetch_array($result);
$mydata=$row[0];
$unser=unserialize($mydata);

echo "$mydata<br>";
print_r($unser);
echo "<br>";
echo $unser[1901];

The output is this:

a:2:{i:2070;s:4:"0.00";i:1901;s:4:"1.00";}
Array ( [2070] => 0.00 [1901] => 1.00 ) 
1.00

So far, so good. Now, I'm trying to write the code so that it checks if the array key 1901 exists. For that, I tried this:

$search_array = $unser;
if (array_key_exists('1901', $search_array)) {
     echo "The key 1901 is in the array";
}

However that is returning an error. What am I doing wrong?

Of course, sorry! The error is: Parse error: syntax error, unexpected T_VARIABLE — user1227914
– user1227914, Commented Apr 1, 2012 at 19:47
All your semicolons are in the right place? i.e. there is no semicolon missing on the line before the erroneous line? — knittl
– knittl, Commented Apr 1, 2012 at 19:52

Gabriel Santos · Accepted Answer · 2012-04-01 20:06:29Z

4

With the following code:

$mydata= 'a:2:{i:2070;s:4:"0.00";i:1901;s:4:"1.00";}';
$unser=unserialize($mydata);

echo "$mydata<br>";
print_r($unser);
echo "<br>";
echo $unser['1901'];

$search_array = $unser;
if (array_key_exists('1901', $search_array)) {
     echo "<br />The key 1901 is in the array";
}

It will work correctly:

a:2:{i:2070;s:4:"0.00";i:1901;s:4:"1.00";}
Array ( [2070] => 0.00 [1901] => 1.00 )
1.00
The key 1901 is in the array

Check if you have more code after the lines of code you have posted. I think is another piece of code which is confusing you.

edited Apr 1, 2012 at 20:06

answered Apr 1, 2012 at 20:03

Gabriel Santos

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Comments

David · Accepted Answer · 2012-04-01 19:51:36Z

-1

echo $unser[1901];

should be

echo $unser['1901'];

Also you can do

if(isset($unser['1901'])) { }

instead of array_key_esists()

answered Apr 1, 2012 at 19:51

David

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5 Comments

Gabriel Santos Over a year ago

Correct, but not 100%. Some times, PHP recognize you are doing some mistakes and know 1901 is not a constant.

Pateman Over a year ago

$unser[1901] is completely fine. And array_key_exists() and isset() are not the same in this case.

user1227914 Over a year ago

That didn't work unfortunately. Now the error is Parse error: syntax error, unexpected T_CONSTANT_ENCAPSED_STRING

David Over a year ago

Post more source code. Error is probably happening 1 line before posted code.

user1227914 Over a year ago

It's strange that's all the code there is between the echo that works fine and the search_array I initially tried

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