PHP Loop not working for CSV output using MySQL variable

Question

I created a variable from a MySQL query that gives a total number value of photos available for a record. This along with a few other variables from the query I can use to construct a dynamic URL in a CSV file I am exporting.

This works great:

  foreach($rowr as $name => $value)

  {
      if($name=='MatrixImage') {$jpg_name=$value;}
      if($name=='PhotoCount') {$count_name=$value;}     
      if($name != 'estate_property_gallery')
            $csv_output .= $value."|";


      if($name == 'estate_property_gallery')
    {
      $csv_output .= "http://url/feeds/".$value."/rets_images/".$jpg_name."_".$count_name.".jpg";
    }               

  }

I need to be able to loop through and create multiple URLs using this PhotoCount variable starting from 1 to the final value found in PhotoCount seperated by a comma but I can't seem to get it right.

Not working:

  foreach($rowr as $name => $value)

  {
      if($name=='MatrixImage') {$jpg_name=$value;}
      if($name=='PhotoCount') {$count_name=$value;}     
      if($name != 'estate_property_gallery')
            $csv_output .= $value."|";


      if($name == 'estate_property_gallery')
    {
      //$csv_output .= "http://url/feeds/".$value."/rets_images/".$jpg_name."_".$count_name.".jpg";
      for( $pic_start=1; $pic_start<=$count_name; $pic_start++ ){
    echo '$csv_output .= "http://url/feeds/' . $value . '/rets_images/' . $jpg_name . '_' . $pic_start . '.jpg,"';
}
    }               

  }

The example above outputs the entire line including the $csv_output to the screen instead of putting each URL into the column of the CSV and delimiting with a comma.

To clarify: I need to be able to write this echo into the $csv_output .= portion of the code successfully so I can write multiple URLs separated by a comma within that column in my csv output:

echo 'http://www.myagentsbuddy.com/feeds/' . $value . '/rets_images/' . $jpg_name . '_' . $count_name . '.jpg';

this => echo '$csv_output single quotes don't get parsed. Then a comma in '.jpg," if that is intentional or not, no idea. Then closing in a single quote. Fix that, code should work. — Funk Forty Niner
– Funk Forty Niner, Commented Aug 11, 2015 at 11:47
so basically echo $csv_output .= "http://url/feeds/' . $value . '/rets_images/' . $jpg_name . '_' . $pic_start . '.jpg,"; — Funk Forty Niner
– Funk Forty Niner, Commented Aug 11, 2015 at 11:52
@Fred-ii- still outputs to screen when I update it with this: echo "$csv_output .= 'url' . $value . '/rets_images/' . $jpg_name . '_' . $pic_start . '.jpg|';"; — Rocco The Taco
– Rocco The Taco, Commented Aug 11, 2015 at 12:16
if echo "$csv_output .= 'url' . $value . '/rets_images/' . $jpg_name . '_' . $pic_start . '.jpg|';"; there's an extra ; in there after '. - but you say it outputs to screen, what exactly do you mean by that? — Funk Forty Niner
– Funk Forty Niner, Commented Aug 11, 2015 at 12:17
output to screen shows a mangled column too instead of a URLs delimited by a comma. — Rocco The Taco
– Rocco The Taco, Commented Aug 11, 2015 at 12:18

Don't Panic · Accepted Answer · 2015-08-11 13:49:51Z

1

If the first example you showed works great, and does not include an echo, and the second example you showed (with an echo) does not work, it seems like it could be fixed by removing the echo and going back to just appending the URLs to the $csv_output like you're doing in the working example.

if($name == 'estate_property_gallery') {
    $comma = '';
    for( $pic_start=1; $pic_start<=$count_name; $pic_start++ ) {
        $csv_output .= $comma.'http://url/feeds/'.$value.'/rets_images/'
                      .$jpg_name.'_'.$pic_start.'.jpg';
        // doing the comma like this will prevent a trailing comma after the last entry
        $comma = ',';
    }
}

answered Aug 11, 2015 at 13:49

Don't Panic

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1 Comment

Rocco The Taco Over a year ago

Brilliant! Thanks so much for the help, basically looks like I over-complicated it LOL!

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