Insert array contents into a string

Question

I am trying to insert the contents of an array in to a string using PHP. My array ($array1) looks like this:

Array1  
    (  
       [0] => http://www.example.com/1  
       [1] => http://www.example.com/2  

    )

I want to insert both links in to a coma separated string, so I can then insert it in to a database field.

I tried this:

foreach ($array1 as $name => $value) {
          $string1 .= $value . ",";
          }
  echo $string1;

Which does work, but I am doing this twice in my code for another array that I also want in a separate string ($string2)

    Array2  
    (  
       [0] => http://www.example.com/3  
       [1] => http://www.example.com/4  
    )

When I echo $string1 I get the correct output
http://www.example.com/1,http://www.example.com/2

But $string2 becomes this:
http://www.example.com/1,http://www.example.com/2,http://www.example.com/3,http://www.example.com/4

This happens even if I use different variable names in the foreach loop above.

Someone else also suggested I try this:

$string1 = implode(',' , $array1);

But I'm not getting any output.

Any help as to how to solve this, or any different approach is greatly appreciated!

To get ouptut, you need to echo, print, etc, $string1 (see my answer below). The code you show, just sets $string1 to a certain value, but it doesn't display that value. — Peter Ajtai
– Peter Ajtai, Commented Jul 21, 2010 at 2:30
Yes, I was echoing it but got no output because I was using the wrong variable name :/ I need to get off the computer for a while.Thanks! — RafaelM
– RafaelM, Commented Jul 21, 2010 at 2:32

Jacob Relkin · Accepted Answer · 2010-07-21 02:11:00Z

2

There's a PHP function called implode for this exact purpose.

$csv = implode(',', $array);

echo $csv; //blah,blah,blah,blah

answered Jul 21, 2010 at 2:11

Jacob Relkin

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2 Comments

Dagg Nabbit Over a year ago

Implode, mentioned in OP. Looks like he expected it to output something.

Dagg Nabbit Over a year ago

heh, this said explode before. Knew I should have kept quiet for 5 more minutes ;)

Dagg Nabbit · Accepted Answer · 2010-07-21 02:11:04Z

1

implode should work fine. It's won't give you any output unless you echo or otherwise output the result, of course.

answered Jul 21, 2010 at 2:11

Dagg Nabbit

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4 Comments

RafaelM Over a year ago

I'm echoing of course after the implode, still no output though

RafaelM Over a year ago

argh just noticed why I got no output after implode. I was using the wrong variable name. Thanks!

Dagg Nabbit Over a year ago

Heh. Posting code helps... just make sure you read it while you're posting; you may find you don't need to ask anything after getting a fresh look at it. Sounds silly but it's happened to me more than once, now I just paste a scaled down copy into another buffer and take a fresh look at it.

RafaelM Over a year ago

Yes, this is a small script I'm writing to learn programming. Extracting the code from the rest of the script definitely helps. Also, getting up for a while from the computer helps too hehe. Thanks!

Kristoffer Bohmann · Accepted Answer · 2011-11-15 16:17:17Z

0

$array1 = array("http://www.example.com/1", "http://www.example.com/2");
$array2 = array("http://www.example.com/3", "http://www.example.com/4");
echo implode(", ", array_merge($array1,$array2));

Output:

http://www.example.com/1, http://www.example.com/2, http://www.example.com/3, http://www.example.com/4

answered Nov 15, 2011 at 16:17

Kristoffer Bohmann

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