php file_exists does not work for array

Question

I posted this question earlier but I have now used the feedback and simplified the php program to show how it still fails. using the file_exists with an array always fails: Here is a simple program I wrote that shows the failure:

[root@dsmpp1 steve]# ls -l data
 total 4 -rw-r--r-- 1 root root 0 Sep 19 11:41 test_file 
[root@dsmpp1 steve]# cat test_file.php
`#!/usr/bin/php -q 
    <?php 
    $i=1; 
    $testarray=array(); 
    $testarray[$i]="test_file"; 
    echo "testarray $testarray[$i]\n";
     **if(file_exists("/home/steve/data/testarray[$i]")) {**
    echo "file exists\n"; } 
    else { echo "file does not exist\n"; } `    
[root@dsmpp1 steve]# php -q test_file.php 
testarray test_file 
file does not exist 
[root@dsmpp1 steve]#

I used the double quotes around the directory and file name as suggested earlier and it is still not working.

Thank you everyone for commenting. Here is a simple program anyone can try and you will get the same results I am getting. Every /etc/hosts file has a "localhost" entry for ip 127.0.0.1. Try populating an array grepping on the IP. The file_exists will not work. Then try hard coding it and see the results. Why when I poplulate an array using a command that returns the same value does the file_exists not work. Please try the following with and without the comment on each of the assignments. — user1683269
– user1683269, Commented Sep 19, 2012 at 20:00
#!/usr/bin/php -q <?php $i=1; $hostname=array(); $hostname[$i]=grep 127.0.0.1 /etc/hosts |awk '{print $3}'; #$hostname[$i]="localhost"; echo "hostname[$i] = $hostname[$i]\n"; if(file_exists("/home/steve/data/$hostname[$i]")) { echo "do nothing appserver file exists\n"; } else { echo "No file exists create one\n"; $touch=touch /home/steve/data/$hostname[$i]; } ?> — user1683269
– user1683269, Commented Sep 19, 2012 at 20:01

user399666 · Accepted Answer · 2012-09-19 16:02:38Z

1

Shouldn't it be:

$testarray[$i]="test_file.php";

instead of:

$testarray[$i]="test_file";

answered Sep 19, 2012 at 16:02

user399666

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2 Comments

user1683269 Over a year ago

no i am not testing for the php script file, I am testing for the file in /home/steve/data/test_file.

DiverseAndRemote.com Over a year ago

The op showed output of ls command showing that the file was named "test_file"

DiverseAndRemote.com · Accepted Answer · 2012-09-19 16:22:01Z

1

try

if(file_exists("/home/steve/data/{$testarray[$i]}")) {**

You were missing the $ before testarray

You might also need to wrap this in brackets because you are using two variables. so use {$testarray[$i]}

edited Sep 19, 2012 at 16:22

answered Sep 19, 2012 at 16:03

DiverseAndRemote.com

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1 Comment

user1683269 Over a year ago

Yes thank you and after adding the $ it did not make a difference. For some reason it is how the array is getting populated. When hard coding the name in the file exists it works fine. As far as I can see the array value has to have something different when I populate it using the following $appserver[$i]=grep $ip[2] /etc/hosts |awk '{print $2}'; If I assign the same name to appserver[$i] by actually assigning the value the file_exists works.

Horen · Accepted Answer · 2012-09-19 16:03:40Z

0

You are missing a $ before testarray in the if clause. try this:

if(file_exists("/home/steve/data/".$testarray[$i])) {

answered Sep 19, 2012 at 16:03

Horen

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5 Comments

Horen Over a year ago

well, does the file actually exist? and what are the ** in front of the if?

Horen Over a year ago

what happens when you hard code the file name in the if condition?

user1683269 Over a year ago

First yes I realized I did not have the $ in front of the testarray.

user1683269 Over a year ago

However when I added the $ it still made no difference. Hard coding the name does work. I do not know why when I populate the array with $appserver[$i]=grep $ip[2] /etc/hosts |awk '{print $2}'; it does not work. Both hardcoding and the above code both yield the same name.

user1683269 Over a year ago

So I have determined it is not the double quotes, or a . before the variable or anything else. It is how the array is being populated.

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