我有一个字符串:<!DOCTYPE html>
<head>
<meta charset="utf-8">
<meta name="viewport" content="width=device-width, initial-scale=1">
<script src="https://code.jquery.com/jquery-3.3.1.min.js" integrity="sha256-FgpCb/KJQlLNfOu91ta32o/NMZxltwRo8QtmkMRdAu8=" crossorigin="anonymous"></script>
<link rel="stylesheet" href="https://maxcdn.bootstrapcdn.com/bootstrap/3.3.7/css/bootstrap.min.css" integrity="sha384-BVYiiSIFeK1dGmJRAkycuHAHRg32OmUcww7on3RYdg4Va+PmSTsz/K68vbdEjh4u" crossorigin="anonymous">
</head>
<body>
<div class="container">
<div class="col-xs-6">
<div class="col-xs-12 col-sm-6">
<div class="col-xs-12">A</div>
<div class="col-xs-12">E</div>
<div class="col-xs-12">I</div>
<div class="col-xs-12">M</div>
</div>
<div class="col-xs-12 col-sm-6">
<div class="col-xs-12">B</div>
<div class="col-xs-12">F</div>
<div class="col-xs-12">J</div>
<div class="col-xs-12">N</div>
</div>
</div>
<div class="col-xs-6">
<div class="col-xs-12 col-sm-6">
<div class="col-xs-12">C</div>
<div class="col-xs-12">G</div>
<div class="col-xs-12">K</div>
<div class="col-xs-12">O</div>
</div>
<div class="col-xs-12 col-sm-6">
<div class="col-xs-12">D</div>
<div class="col-xs-12">H</div>
<div class="col-xs-12">L</div>
<div class="col-xs-12">P</div>
</div>
</div>
</div>
</body>
</html>
基本上我想拥有这种结构-
position1, position2
到目前为止,我已经尝试过了。
array(
array('position' => 'position 1'),
array('position' => 'position 2')
)
$positions = explode(',', "position1, position2");
$modPoses = [];
foreach($positions as $pose):
$modPoses['position'] = $pose;
endforeach;
print_r($modPoses);
Output:
如何获得所需的(如上所述)数组结构? 谢谢。
答案 0 :(得分:2)
将两个值分配给同一索引没有任何意义,但是让我给您一个解决方案。如果您使用2d关联数组,则可以通过这种方式完成
$counter = 0; //initialize counter here
foreach($positions as $pose):
$modPoses[$counter]['position'] = $pose;
$counter++;
endforeach;
答案 1 :(得分:1)
我想你想要的是
array(
array('position' => 'position 1'),
array('position' => 'position 2')
)
如果是这样,您可以使用:
array_map(function ($i) { return array('position' => $i); }, explode(',', 'position 1, position 2'))