php get the number of a given instance in an array

Question

Basically, I'm trying to make a table where the color slightly changes for each row.

I'm going to do this by naming each even row class="even" and every odd row class="odd".

In my loop then, I need to figure out if the instance of the array is odd or even.

I can do almost do this by writing:

if (count($row)%2 == 0)  //while $row is an array from a mysql query
    {
    echo "<tr class='even'>";

if (count($row)%2 == 1)
    {
    echo "<tr class='odd'>";
    }

However writing count($row) gives me the count for the whole array, not the number of the instance which indicates where it falls in the whole array. Any ideas on how I could get the number of the instance within the array?

Thanks

JRL · Accepted Answer · 2011-08-13 05:46:50Z

3

Don't use PHP for this. Use CSS:

tr:nth-child(even) {background: #CCC}
tr:nth-child(odd) {background: #FFF}

answered Aug 13, 2011 at 5:46

JRL

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5 Comments

Jared Ng Over a year ago

... which still needs to be generated using PHP. Jeff's content is presumably not static, since he mentions a database query.

JRL Over a year ago

@Jared: His content can be dynamic, that doesn't mean he should use PHP for the style.

Jared Ng Over a year ago

This answer doesn't solve his problem, which is a PHP one, not a CSS one. Besides, nobody said he wasn't using CSS.

Decent Dabbler Over a year ago

According to quirksmode.org/css/contents.html#t38 nth-child is unsupported in quite some browsers (especially IE).

Jeff Over a year ago

@JRL this css features is very nice -- much simpler than stupid class names - thanks

Jared Ng · Accepted Answer · 2011-08-13 05:51:20Z

1

Try something like this:

$i = 0;
while ($i < count($row)) {
  if (($i % 2) == 0) {
    echo "<tr class='even'>";
  } else {
    echo "<tr class='odd'>";
  }
  $i++;
}

edited Aug 13, 2011 at 5:51

answered Aug 13, 2011 at 5:45

Jared Ng

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3 Comments

Josh Over a year ago

Dont forget to increment $i.

Decent Dabbler Over a year ago

@Jared Ng: Your edit will not produce what you expect. You should increment at the end of the while body.

Jared Ng Over a year ago

This should have no problems. Sorry, it's been a while since I've touched PHP.

bfavaretto · Accepted Answer · 2011-08-13 05:52:59Z

1

You could use a counter:

$i = 0   
while($row=mysql_fetch_array($yourquery)) {
   if ($i % 2 == 0) {
      echo "<tr class='even'>";
   } else {
      echo "<tr class='odd'>";
   } 
   $i++;   
}

answered Aug 13, 2011 at 5:52

bfavaretto

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