foreach输出两个数组首先跳过[重复]

问题描述：

Possible Duplicate:
how to skip elements in foreach loop

I have the following foreach:

foreach($documents as $document): 
    print_r($document);
endforeach;

Which outputs the following:

Array
(
    [num] => 2
)
Array
(
    [0] => Array
        (
            [name] => Batman
            [url] => http://batman.com
        )

    [1] => Array
        (
            [name] => Superman
            [url] => http://superman.com
        )

)

The first array conatining [num] => 2, I dont want to use in my foreach when printing the result.

But how do I get rid of that array so it doesn't get printed when I use write print_r($document)?

可能重复： strong>
如何跳过foreach循环中的元素 p> blockquote> \ n

我有以下foreach： p>

foreach（$ documents as $ document）： print_r（$ document）; endforeach; code> pre> 输出以下内容： p> Array （ [num] =＆gt; 2 \ n） Array （ [0] =＆gt;数组（ [名称] =＆gt;蝙蝠侠 [url] =＆gt; http：//batman.com ） [1] =＆gt;数组（ [名称] =＆gt;超人 [url] =＆gt; http：//superman.com ）） code> pre> 第一个包含[num] =＆gt;的数组 2，打印结果时我不想在我的foreach中使用。 p> 但是如何摆脱那个数组，以便在我使用write print_r（$ document）时不打印？ p> div>

答

Keeping with the foreach you can use continue:

$first = true;

foreach($documents as $document) {
    if($first) {
        $first = false;
        continue;
    }

    print_r($document);
}

答

Use a standard for loop with an incrementing index specifier and skip the first element.

for($i = 1; $i < count($documents); $i++) {
  print_r($documents[i]);
}

答

The easiest approach would be to remove the first array completely, however my guess is that you can't do that. No worries - this should have you covered:

for( $i = 1; $i < count($documents); $i++ ):
   print_r($documents[$i]);
endfor;

Edit: I've created a test case for you on Codepad.org.

foreach输出两个数组首先跳过[重复]

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