在PHP中处理变量
我正在尝试使用变量。通过在foreach循环中使用在PHP中处理变量,php,loops,variables,Php,Loops,Variables,我正在尝试使用变量。通过在foreach循环中使用$type,我想得到$week['one']和$month['one']的值 $types = array( 'week', 'month' ); $week = array( 'one' => 1.2, 'two' => 0.13, ); $month = array( 'one' => 1.2,
$type
,我想得到$week['one']
和$month['one']
的值
$types = array(
'week',
'month'
);
$week = array(
'one' => 1.2,
'two' => 0.13,
);
$month = array(
'one' => 1.2,
'two' => 0.13,
);
我尝试过但没有成功的方法:
<?php foreach ($types as $type): ?>
<?= $$type['one']; ?><br />
<?= $$type['two']; ?><br />
<?php endforeach; ?>
<?php foreach ($types as $type): ?>
<?= ${$type}['one']; ?><br />
<?= ${$type}['two']; ?><br />
<?php endforeach; ?>
<?php foreach ($types as $type): ?>
<?= $($type)['one']; ?><br />
<?= $($type)['two']; ?><br />
<?php endforeach; ?>
一切似乎都会导致语法错误。我是否使用了错误的语法?使用此变体:
<?php foreach ($types as $type): ?>
<?= ${$type}['one']; ?><br />
<?= ${$type}['two']; ?><br />
<?php endforeach; ?>
检查这里:。这需要什么PHP版本。这似乎对我不起作用。为了记录在案。。。我不是投反对票的人:)请再次检查,我添加了一个链接。第一个答案只适用于>=7。事实上,你的变体是正确的。非常抱歉。我有一个打字错误:)我用了
${$type}'week']