使fetch数组包含变量的麻烦

Question

我正在尝试循环数据库例程，但是我在使获取数组包含变量时遇到问题。

${"db" . $currentTime . "11"} = ${"data" . $currentTime . "1"}[${$currentTLower . "11"}];

$currentTime包含'A'

$currentTLower包含'a'

因此代码行如下所示：

$dbA11 = $dataA1[a11];

但是，它没有成功。

Answer 1

你非常接近。索引应为“a11”，但不应与代码

中的$a11相同

$currentTime = 'A';
$currentTLower ='a';
$dataA1['a11'] = 'value';

${"db" . $currentTime . "11"} = ${"data" . $currentTime . "1"}[$currentTLower . "11"];

echo $dbA11; // value

Answer 2

<?php
$currentTime = "A";
$currentTLower = "A1";
$A111 = "Test";
$dataA1[$A111] = "This is content";
${"db".$currentTime."11"} = ${"data".$currentTime . "1"}[${$currentTLower . "11"}];
echo ${"db".$currentTime."11"};

// IT will print : This is content
?>