Laravel explode() url 查询

Question

在我的数据库中，查询输入存储为 BLOB，如下所示： make=ACURA&model%5B0%5D=MDX&model%5B1%5D=NSX&auction%5B0%5D=BAYAUC&auction%5B1%5D=IAA+Osaka

现在在我看来，我试着去搞爆：

@foreach(explode('=', $url) as $par)
     {{ $par }} 
@endforeach

但是我得到一个错误：

htmlspecialchars() expects parameter 1 to be string, object given

我想输出的最终结果是这样的：讴歌 MDX NSX BAYAUC IAA+大阪

Answer 1

我不明白你想做什么，但你用错了 explode() 功能。

正确的方法是array explode ( string $delimiter , string $string [, int $limit = PHP_INT_MAX ] )

The 3rd parameter is limit and MUST BE AN INTEGER which mean If limit is set and positive, the returned array will contain a maximum of limit elements with the last element containing the rest of string.

If the limit parameter is negative, all components except the last -limit are returned.

If the limit parameter is zero, then this is treated as 1.

http://php.net/manual/en/function.explode.php

Answer 2

您可以使用此 php 方法进行 url 查询字符串解析：parse_str($string, $output_array);

PHP

$str = 'make=ACURA&model%5B0%5D=MDX&model%5B1%5D=NSX&auction%5B0%5D=BAYAUC&auction%5B1%5D=IAA+Osaka';
parse_str($str, $output);

// var_dump($output);

查看输出：https://3v4l.org/LLvAV

Laravel Blade 示例：

@foreach( $output as $opt )
    @if( is_array($opt) )
        {{ join(" ", $opt)." " }}
    @else
        {{ $opt." " }}
    @endif
@endforeach

Laravel explode() url 查询

Laravel explode() url query

php

laravel-5.4

PHP

查看输出：https://3v4l.org/LLvAV

Laravel Blade 示例：