Return value from array_push() is an integer instead of the populated array

Question

My code is as follows:

foreach ($row_result as $results)
{
  $f = fopen("jigsaw.csv", "a+");
  $new_array = array();
  $content = $results->findElement(WebDriverBy::className('seo-company'))->geTtext();
  $test = array();
  if ($i == 1)
  {
    $Orgnisation_name = $content;
    $concate = array_push($test,$Orgnisation_name);
    echo "\n";
  }  
  if ($i == 2)
  {
    echo $Website = $content;
    echo "\n";
  }  
  if ($i == 3)
  {
    echo $HeadQuarters = $content;
    echo "\n";
  }  
  if ($i == 4)
  {
    echo $Phone = $content;
    echo "\n";
  }  
  if ($i == 5)
  {
    echo $Industries = $content;
    echo "\n";
  }  
  if ($i == 6)
  {
    echo $Employees = $content;
    echo "\n";
  }  
  if ($i == 7)
  {
    echo $Revenue = $content;
    echo "\n";
  }  
  if ($i == 8)
  {
    echo $Ownership = $content;
    echo "\n";
  }  
  $i++;
}

Here I want to push one by one element into concate array and finally save it into mysql php database.

When I try to push it into concate array its printing me 1 which is wrong.

What should I do?

Viacheslav Kondratiuk · Accepted Answer · 2013-11-30 12:20:31Z

2

array_push return value is:

Returns the new number of elements in the array.

So 1 is the number of elements in array.

Seems that you will always have 1, because you call $i++ and according to your code, block with array_push will be executed once.

answered Nov 30, 2013 at 12:20

Viacheslav Kondratiuk

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Comments

aland · Accepted Answer · 2013-11-30 12:36:05Z

1

Do you need to use array_push? $concate = array_push($test,$Orgnisation_name); $test is blank at this point. If you just want to add an item to the $concate array, $concate[] = $Orgnisation_name; is enough.

edited Nov 30, 2013 at 12:36

answered Nov 30, 2013 at 12:21

aland

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