如何在Javascript中访问PHP数组?

时间:2022-05-16 08:57:19

I have a very simple PHP array

我有一个非常简单的PHP数组

$array = [];
$array['a'] = '1';
$array['b'] = '2';
$array['c'] = '3';

PHP

If I dd($array); out I got

如果我dd($ array);我得到了

array:3 [▼
  "a" => "1"
  "b" => "2"
  "c" => "3"
]

If I decode dd(json_encode($array));, I got this

如果我解码dd(json_encode($ array));,我得到了这个

"{"a":"1","b":"2","c":"3"}"

JS

I want to be able to access this variable in my Javascript, So I've tried

我希望能够在我的Javascript中访问此变量,所以我试过了


1

console.log($array);

I got

$array is not defined

$ array未定义


2

I'm using Laravel. {{ }} == echo

我正在使用Laravel。 {{}} == echo

console.log('{{$array}}');

I got

500 Internal Error

500内部错误

htmlentities() expects parameter 1 to be string, array given (View: /Users/bheng/Sites/portal/resources/views/cpe/index.blade.php)

htmlentities()期望参数1为字符串,给定数组(查看:/Users/bheng/Sites/portal/resources/views/cpe/index.blade.php)


3

console.log('{{ json_encode($array)}}');

I got

The page to load, but the data is very bad looking

要加载的页面,但数据看起来很糟糕

{"a":"1","b":"2","c":"3"}


4

console.log(JSON.parse('{{ json_encode($array)}}'));

I got

Uncaught SyntaxError: Unexpected token & in JSON at position 1

未捕获的SyntaxError:位于1的JSON中出现意外的令牌


5

console.log(JSON.parse('{{ json_decode($array)}}'));

I got

json_decode() expects parameter 1 to be string, array given

json_decode()期望参数1为字符串,给定数组


6

console.log('{{ json_decode($array)}}');

I got

json_decode() expects parameter 1 to be string, array given

json_decode()期望参数1为字符串,给定数组


GOAL

I just want to be able to access my array as Javascript Array or JSON in the Javascript.

我只是希望能够在Javascript中以Javascript Array或JSON的形式访问我的数组。

Can someone please fill me in on this ?

有人可以请我填一下吗?

4 个解决方案

#1


6  

In Blade, {{ $variable }} will output an escaped version of the string, passed through htmlentities() to make it safe for use in HTML. You want an unescaped version. You can use {!! $variable !!} for that:

在Blade中,{{$ variable}}将输出字符串的转义版本,通过htmlentities()传递,以便在HTML中使用。你想要一个未转义的版本。您可以使用 {!! $ variable !!}为此:

console.log({!! json_encode($array) !!});

You don't need to add quotes around it, json_encode() outputs a valid javascript object. It will add quotes where necessary, if you add them yourself you will get the JSON string in your javascript, instead of the JSON object.

你不需要在它周围添加引号,json_encode()输出一个有效的javascript对象。它会在必要时添加引号,如果你自己添加它们,你将获得javascript中的JSON字符串,而不是JSON对象。

#2


1  

In Laravel you can use {!! !!} to skip entity escaping

在Laravel你可以使用{!! !!}跳过实体转义

console.log({!! json_encode($array) !!});

#3


0  

Just echo it as json data and use it in javascript.

只需将其作为json数据回显并在javascript中使用它。

<?php
  $array = [];
  $array['a'] = '1';
  $array['b'] = '2';
  $array['c'] = '3';
?>
<script>var jsArr = <?=json_encode($array);?>;
alert(jsArr);</script>

EDIT because of clarification that you're using blade. Then it should be:

编辑,因为澄清你正在使用刀片。那应该是:

<?php
  $array = [];
  $array['a'] = '1';
  $array['b'] = '2';
  $array['c'] = '3';
?>
<script>var jsArr = {!! json_encode($array) !!};
alert(jsArr);</script>

{ ... } is an escaped version of your string. But you need the unescapt string. This can be achieved by using {!! ... !!}.

{...}是字符串的转义版本。但是你需要unescapt字符串。这可以通过使用{!! ...... !!}。

#4


0  

First, you have to understand that PHP run on server side and javascript on client side, as PHP make the response you should print a script like this:

首先,您必须了解PHP在服务器端运行并在客户端运行javascript,因为PHP会做出响应,您应该打印如下脚本:

echo "<script>
var sheison = JSON.parse(".dd(json_encode($array)).");
console.log(sheison);
</script>";

I didn't test the code, is just the idea.

我没有测试代码,只是想法。

#1


6  

In Blade, {{ $variable }} will output an escaped version of the string, passed through htmlentities() to make it safe for use in HTML. You want an unescaped version. You can use {!! $variable !!} for that:

在Blade中,{{$ variable}}将输出字符串的转义版本,通过htmlentities()传递,以便在HTML中使用。你想要一个未转义的版本。您可以使用 {!! $ variable !!}为此:

console.log({!! json_encode($array) !!});

You don't need to add quotes around it, json_encode() outputs a valid javascript object. It will add quotes where necessary, if you add them yourself you will get the JSON string in your javascript, instead of the JSON object.

你不需要在它周围添加引号,json_encode()输出一个有效的javascript对象。它会在必要时添加引号,如果你自己添加它们,你将获得javascript中的JSON字符串,而不是JSON对象。

#2


1  

In Laravel you can use {!! !!} to skip entity escaping

在Laravel你可以使用{!! !!}跳过实体转义

console.log({!! json_encode($array) !!});

#3


0  

Just echo it as json data and use it in javascript.

只需将其作为json数据回显并在javascript中使用它。

<?php
  $array = [];
  $array['a'] = '1';
  $array['b'] = '2';
  $array['c'] = '3';
?>
<script>var jsArr = <?=json_encode($array);?>;
alert(jsArr);</script>

EDIT because of clarification that you're using blade. Then it should be:

编辑,因为澄清你正在使用刀片。那应该是:

<?php
  $array = [];
  $array['a'] = '1';
  $array['b'] = '2';
  $array['c'] = '3';
?>
<script>var jsArr = {!! json_encode($array) !!};
alert(jsArr);</script>

{ ... } is an escaped version of your string. But you need the unescapt string. This can be achieved by using {!! ... !!}.

{...}是字符串的转义版本。但是你需要unescapt字符串。这可以通过使用{!! ...... !!}。

#4


0  

First, you have to understand that PHP run on server side and javascript on client side, as PHP make the response you should print a script like this:

首先,您必须了解PHP在服务器端运行并在客户端运行javascript,因为PHP会做出响应,您应该打印如下脚本:

echo "<script>
var sheison = JSON.parse(".dd(json_encode($array)).");
console.log(sheison);
</script>";

I didn't test the code, is just the idea.

我没有测试代码,只是想法。