Array Help --> Find matching ones?

jjfletch

Hi there,

I think this might be difficult, or, dare I say, impossible?

If, for example, the stuff below were my arrays, is there a way to find which ones have a matching [0] value and find the total count and also return it's value?

For example:

Let's say my sql table has two fields, name and groceries, where groceries are imploded upon entry and exploded upon retrieval.

Ultimately, the values of $example (in various cases) looks like this:

$example = array("fruits", "vegetables", "meat");
$example = array("vegetables", "milk", "cheese", "bread", "eggs");
$example = array("fruits", "milk", "nuts");
$example = array("fruits", "cheese", "bread");
$example = array("vegatables", "milk", "cheese", "bread");

What I want to do is:

(1) Find which arrays have a matching [0] element. In this case, cases 1, 3, and 4 match, and cases 2 and 5 match.

(2) Then, I want to print something like this:

Fruits 3
Vegetables 2

I should also mention that I don't know the string values that begin the records. For all I know, it could be eggs or bread that starts it off. So, I don't want to do something that would require having to search for a string within the array.

Is this even possible? Can someone help me out here?

Thanks!

bpat1434

It would probably have something that looks like this:

<?php

$example = array("fruits", "vegetables", "meat");
$example1 = array("vegetables", "milk", "cheese", "bread", "eggs");
$example2 = array("fruits", "milk", "nuts");
$example3 = array("fruits", "cheese", "bread");
$example4 = array("vegatables", "milk", "cheese", "bread");

$count = array(); // Will hold our count values

if($example[0] == $example1[0])
    $count[$example[0]] = ($count[$example[0]] != '')?$count[$example[0]++:0;
if($example[0] == $example2[0])
    $count[$example[0]] = ($count[$example[0]] != '')?$count[$example[0]++:0;
// etc...etc...
?>

It's a basic if statement. Now, if you had all that info in one array, like:

Array(
    [0] => array(
        [0] => 'fruits',
        [1] => 'vegetables',
        [2] => 'meat'
    ),
    [1] => Array(
        [0] => 'vegetables',
        [1] => 'milk',
        [2] => 'cheese',
        [3] => 'bread',
        [4] => 'eggs'
    ),
    [2] => Array(
        [0] => 'fruits',
        [1] => 'milk',
        [2] => 'nuts'
    ),
    [3] => Array(
        [0] => 'fruits',
        [1] => 'cheese',
        [2] => 'bread'
    ),
    [4] => Array(
        [0] => 'vegetables',
        [1] => 'milk',
        [2] => 'cheese'
        [3] => 'bread',
    )
)

You could just itterate through the main array and compare... or itterate and do an array search, and if the keys match increase the count.

Now, why is it crucial that the keys match?

jjfletch

Hmm, going with the arrays within a main array, what would that look like?

I would be able to build that just by doing:

$bigArray = array($example);

Like that? And then how would I just iterate through them?

Thanks for your help! It's starting to make sense....

bpat1434

$count = array();  // Our counter!!
foreach($array as $home=>$newarray)
{
  foreach($newarray as $key=>$val)
  {
    foreach($array as $key2=>$val2)
    {
      if($key != $key2) // Not the same array we're looking at
      {
        if(false != ($key=array_search($val, $array[$key2]))
        {
          // Add incrementation just like my last post
        }
      }
    }
  }
}

jjfletch

Sorry, but I'm just getting a parse error: Parse error: parse error, unexpected '{' ... on Line 59

Here's my code with an attempt at implementing yours:

while($row = mysql_fetch_assoc($result)) {

    $vs = explode(",", $row['example']);

    $bigArray1 = array($vs);

$count = array(); 
foreach($array as $home=>$newarray)
{ 
  foreach($newarray as $key=>$val)
  { 
    foreach($array as $key2=>$val2)
    { 
      if($key != $key2)
      { 
        if(false != ($key=array_search($val, $array[$key2]))  

        { /******* THIS IS LINE 59 ********/
        $count[$vs[0]] = ($count[$vs[0]] != '')?$count[$vs[0]++:0;      

        }
      }
    }
  }
} 

}

bradgrafelman

Missing one ')' on the preceding line:

        if(false != ($key=array_search($val, $array[$key2]))

should be:

        if(false != ($key=array_search($val, $array[$key2])))

jjfletch

Thanks for the guidance, but I'm still getting errors. Now I'm getting invalid argument errors. Here it is again:

$query = "SELECT * FROM myTable WHERE example like '%eggs%' OR example like '%milk%' AND example not like '%eggs%'";
$result = mysql_query($query);

while($row = mysql_fetch_assoc($result)) {

    $vs = explode(",", $row['example']);   

    $bigArray1 = array($vs);   

$count = array();
foreach($bigArray1 as $home=>$newarray)
{ 
  foreach($newarray as $key=>$val)
  {
    foreach($array as $key2=>$val2)
    {
      if($key != $key2)
      {
        if(false != ($key=array_search($val, $array[$key2])))
        {                                   

        $count[$vs[0]] = ($count[$vs[0]] != '')?$count[$vs[0]]++:0;
        }
      }
    }
  }
} 

}

bradgrafelman

If you're getting PHP errors, post the exact error message PHP is giving you.

jjfletch

Warning: Invalid argument supplied for foreach() in ... on line 51
Warning: Invalid argument supplied for foreach() in ... on line 51

bradgrafelman

I think your problem is this:

        $bigArray1 = array($vs);    

$count = array(); 
foreach($bigArray1 as $home=>$newarray)

What this does is take the array $vs and sticks it inside $bigArray1 in key 0. So in order to get the $vs array back, you actually want $bigArray1[0]. Why not just change that bit of code to this:

$count = array(); 
foreach($vs as $home=>$newarray)

Also, looking ahead, this line:

$count[$vs[0]] = ($count[$vs[0]] != '')?$count[$vs[0]]++:0;

is going to cause problems. The '++' tells PHP to increase the value of the variable by one. It does not say "return the value of one more than the variable". I believe you'll want to change that to this:

$count[$vs[0]] = (($count[$vs[0]] != '') ? $count[$vs[0]]+1 : 0);

I added a second set of ()'s simply because I've always done that with ternary statements to make sure PHP evaluates them correctly. The added whitespace is simply for readability purposes.

jjfletch

🙁

I'm still getting:

Warning: Invalid argument supplied for foreach() in ... on line 51

Line 51 is: foreach($newarray as $key=>$val)

bradgrafelman

That means $newarray isn't an array.

I think you need to use [man]print_r/man to print out your array, and look at your logic. I don't think you need that many foreach() statements.