使fetch数组包含变量的问题 - Trouble with making fetch array contain variable

Trouble with making fetch array contain variable

本文关键字：问题变量包含 fetch 数组 | 更新日期: 2023-09-26

我正在尝试循环数据库例程，但我有问题，使fetch数组包含一个变量。g

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

$currentTime包含'A'

$currentTLower包含'a'

那么这行代码应该是这样的:

$dbA11 = $dataA1[a11];

但是，它不起作用。

你很接近了。索引应该是'a11'，而不是$a11在你的代码

$currentTime = 'A';
$currentTLower ='a';
$dataA1['a11'] = 'value';
${"db" . $currentTime . "11"} = ${"data" . $currentTime . "1"}[$currentTLower . "11"];
echo $dbA11; // value

<?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
?>