I'm scraping a website using php to get some data. The data I get is a valid javascript array.
我正在使用php抓取一个网站来获取一些数据。我得到的数据是一个有效的javascript数组。
"['v42, 2015', 23428, 30243, 76993]
,
['v43, 2015', 24060, 30401, 73412]
,
['v44, 2015', 22855, 29720, 71573]
,
['v45, 2015', 24455, 30757, 78991]
,
['v46, 2015', 24275, 30398, 84424]"
I now have this string in php, but how can I convert it to a php array?
我现在在php中有这个字符串,但是如何将其转换为php数组呢?
3 个解决方案
#1
6
$string = "['v42, 2015', 23428, 30243, 76993]
,
['v43, 2015', 24060, 30401, 73412]
,
['v44, 2015', 22855, 29720, 71573]
,
['v45, 2015', 24455, 30757, 78991]
,
['v46, 2015', 24275, 30398, 84424]";
It is a valid js array if you add the proper start & end square brackets delimiter. Furthermore, to comply with the php json parser requirements, the string delimiters must be double-quoted instead of single-quoted, so a quick replacement must be done.
如果添加正确的开始和结束方括号分隔符,它是一个有效的js数组。此外,为了符合php json解析器要求,字符串分隔符必须是双引号而不是单引号,因此必须快速替换。
You then can decode it like so :
然后您可以像这样解码它:
$ary = json_decode('['.str_replace("'",'"',$string).']', true);
#2
2
The single quotes may be valid in JS, but JSON sometimes have a problem with it. You can try it here: JSONLint
单引号可能在JS中有效,但JSON有时会遇到问题。你可以在这里试试:JSONLint
To get a valid JSON, just replace the single quotes '
with double quotes "
, to get an array with arrays you have to surround your string with brackets []
.
要获得有效的JSON,只需将单引号'替换为'双引号',以获得具有数组的数组,您必须用括号[]包围字符串。
Try this example code:
试试这个示例代码:
$string = "['v42, 2015', 23428, 30243, 76993]
,
['v43, 2015', 24060, 30401, 73412]
,
['v44, 2015', 22855, 29720, 71573]
,
['v45, 2015', 24455, 30757, 78991]
,
['v46, 2015', 24275, 30398, 84424]";
$string = str_replace( "'" , '"', $string );
$string = '['.$string.']';
echo "<pre>";
var_dump( json_decode( $string ) );
#3
1
You can try replacing the [ ]
with ''
and then breaking the string.
你可以尝试用''替换[]然后打破字符串。
$string = str_replace(']', '', str_replace('[', '',$string));
$array = explode(',', $string);
#1
6
$string = "['v42, 2015', 23428, 30243, 76993]
,
['v43, 2015', 24060, 30401, 73412]
,
['v44, 2015', 22855, 29720, 71573]
,
['v45, 2015', 24455, 30757, 78991]
,
['v46, 2015', 24275, 30398, 84424]";
It is a valid js array if you add the proper start & end square brackets delimiter. Furthermore, to comply with the php json parser requirements, the string delimiters must be double-quoted instead of single-quoted, so a quick replacement must be done.
如果添加正确的开始和结束方括号分隔符,它是一个有效的js数组。此外,为了符合php json解析器要求,字符串分隔符必须是双引号而不是单引号,因此必须快速替换。
You then can decode it like so :
然后您可以像这样解码它:
$ary = json_decode('['.str_replace("'",'"',$string).']', true);
#2
2
The single quotes may be valid in JS, but JSON sometimes have a problem with it. You can try it here: JSONLint
单引号可能在JS中有效,但JSON有时会遇到问题。你可以在这里试试:JSONLint
To get a valid JSON, just replace the single quotes '
with double quotes "
, to get an array with arrays you have to surround your string with brackets []
.
要获得有效的JSON,只需将单引号'替换为'双引号',以获得具有数组的数组,您必须用括号[]包围字符串。
Try this example code:
试试这个示例代码:
$string = "['v42, 2015', 23428, 30243, 76993]
,
['v43, 2015', 24060, 30401, 73412]
,
['v44, 2015', 22855, 29720, 71573]
,
['v45, 2015', 24455, 30757, 78991]
,
['v46, 2015', 24275, 30398, 84424]";
$string = str_replace( "'" , '"', $string );
$string = '['.$string.']';
echo "<pre>";
var_dump( json_decode( $string ) );
#3
1
You can try replacing the [ ]
with ''
and then breaking the string.
你可以尝试用''替换[]然后打破字符串。
$string = str_replace(']', '', str_replace('[', '',$string));
$array = explode(',', $string);