哪个 php 配置变量不允许连接数组索引？

发布于 2024-11-19 18:02:31 字数 475 浏览 9 评论 0原文

if(!isset($am_states[$lot.'_-40C'])){

或

$am_states[$temp."_".$states[$i]['temperature']] = $states[$i]['temperature'];

每当我有带有连接字符串作为数组键的数组时，php 都会返回错误：

解析错误：语法错误，意外的“.”，期望“]”

所以我假设服务器出现问题配置，尽管我确信我更改了本地配置的某些内容。上次我更改配置是在设置 apache/mysql/php 安装时

，即 PHP 版本 5.3.1、Apache/2.2.14、MYSQL5.1.41（默认来自 xampp1.7.3），

所以我在 7 个月前使用了此语法他们工作正常。只是现在他们产生了错误。有人可以帮忙吗？

原文

E.G.

if(!isset($am_states[$lot.'_-40C'])){

$am_states[$temp."_".$states[$i]['temperature']] = $states[$i]['temperature'];

Whenever I have arrays with concatenated string as array-keys php returns an error:

Parse error: syntax error, unexpected '.', expecting ']'

So I am assuming something is wrong with the server configuration although I am sure i changed something on my local configuration.
Last time i changed the configuration was when i setup my apache/mysql/php installation

that is PHP Version 5.3.1,Apache/2.2.14,MYSQL5.1.41 (default from xampp1.7.3)

so I was using this syntaxes 7 months ago and they were working properly. It just now that they produce errors.
Anyone can help?

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橙幽之幻 2024-11-26 18:02:31

您对您正在使用的 PHP 版本确定吗？以下测试（使用 PHP 5.3.6 (cli)）可以正常运行。也许您可以发布一个更完整的示例？

#!/usr/bin/env php
<?php

$states = array(
    array('temperature' => 40),
    array('temperature' => 50),
    array('temperature' => 60)
);

$temp = 'test';
$i = 2;

$am_states[$temp . "_" . $states[$i]['temperature']] = $states[$i]['temperature'];

var_dump($am_states);

该脚本的输出是：

array(1) {
  ["test_60"]=>
  int(60)
}

Are you positive about the PHP version you're using? The following test (using PHP 5.3.6 (cli)) works without issue. Perhaps you could post a more complete example?

#!/usr/bin/env php
<?php

$states = array(
    array('temperature' => 40),
    array('temperature' => 50),
    array('temperature' => 60)
);

$temp = 'test';
$i = 2;

$am_states[$temp . "_" . $states[$i]['temperature']] = $states[$i]['temperature'];

var_dump($am_states);

The output of this script is: