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PHP 一个数组操作方法求解解决方案

WBOY
WBOYOriginal
2016-06-13 12:18:32939browse

PHP 一个数组操作方法求解
一维数组

<br />array('0'=>'a',1=>'b',2=>'c',3=>'d',4=>'e',5=>'f');<br />

可能很长

我想转成
<br />array('a'=>'b','c'=>'d','e'=>'f');<br />

这样子,就是第一个值作为key,第二个值作为value,以此类推,求方法!
------解决思路----------------------
楼主你这样做,就要保证你的数据必须是偶数的,不然最后肯定有不完整。当你数据是偶数时,就可以进行遍历
<br /><?php <br />$arr = array('0'=>'a',1=>'b',2=>'c',3=>'d',4=>'e',5=>'f','6'=>'g','7'=>'h');<br />$i = 0;<br />$len = count($arr);<br />$newArr = array();<br />while($i < $len){<br />	$newArr[$arr[$i]] = $arr[$i+1];<br />	$i += 2;<br />}<br />var_dump($newArr);<br />?> <br />

------解决思路----------------------
把楼上的稍微改了下。。 这样应该就可以了 不用管奇偶数了
<br /><?php <br />$arr = array('0'=>'a',1=>'b',2=>'c',3=>'d',4=>'e',5=>'f','6'=>'g','7'=>'h','8'=>'i');<br />$i = 0;<br />$len = count($arr);<br />$newArr = array();<br />while($i < $len){<br />    $newArr[$arr[$i]] = array_key_exists($i+1,$arr)?$arr[$i+1]:null;<br />    $i += 2;<br />}<br />var_dump($newArr);<br />?> <br />

------解决思路----------------------
$a = array('0'=>'a',1=>'b',2=>'c',3=>'d',4=>'e',5=>'f', 'G');<br />foreach(array_chunk($a, 2) as $t) {<br />  $b[$t[0]] = @$t[1];<br />}<br />print_r($b);<br /><br />
Array<br />(<br />    [a] => b<br />    [c] => d<br />    [e] => f<br />    [G] => <br />)<br /><br />

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