0

我正在为我的 PHP 应用程序接收一些结构化数据，但格式有些不可预测且难以处理。我对数据的初始格式没有发言权。我得到的是一个字符串（下面给出的示例）。

[9484,'Víctor Valdés',8,[[['accurate_pass',[15]],['touches',[42]],['saves',[4]],['total_pass',[24]],['good_high_claim',[2]],['formation_place',[1]]]],1,'GK',1,0,0,'GK',31,183,78],[1320,'Carles Puyol',7.76,[[['accurate_pass',[50]],['touches',[75]],['aerial_won',[3]],['total_pass',[55]],['total_tackle',[1]],['formation_place',[6]]]],2,'DC',5,0,0,'D(CLR)',35,178,80],[5780,'Dani Alves',8.21,[[['accurate_pass',[58]],['touches',[99]],['total_scoring_att',[1]],['total_pass',[66]],['total_tackle',[6]],['aerial_lost',[1]],['fouls',[4]],['formation_place',[2]]]],2,'DR',22,0,0,'D(CR)',30,173,64],[83686,'Marc Bartra',8.31,[[['accurate_pass',[64]],['touches',[88]],['won_contest',[1]],['total_scoring_att',[1]],['aerial_won',[1]],['total_pass',[66]],['total_tackle',[5]],['aerial_lost',[1]],['fouls',[1]],['formation_place',[5]]]],2,'DC',15,0,0,'D(C)',22,181,70],[13471,'Adriano',6.72,[[['accurate_pass',[16]],['touches',[28]],['aerial_won',[2]],['total_pass',[18]],['total_tackle',[1]],['formation_place',[3]]]],2,'DL',21,1,31,'D(CLR),M(LR)',29,172,67]

以上是5名足球运动员的数据。这是我需要得到的：

[9484,'Víctor Valdés',8,[[['accurate_pass',[15]],['touches',[42]],['saves',[4]],['total_pass',[24]],['good_high_claim',[2]],['formation_place',[1]]]],1,'GK',1,0,0,'GK',31,183,78]

[1320,'Carles Puyol',7.76,[[['accurate_pass',[50]],['touches',[75]],['aerial_won',[3]],['total_pass',[55]],['total_tackle',[1]],['formation_place',[6]]]],2,'DC',5,0,0,'D(CLR)',35,178,80]

[5780,'Dani Alves',8.21,[[['accurate_pass',[58]],['touches',[99]],['total_scoring_att',[1]],['total_pass',[66]],['total_tackle',[6]],['aerial_lost',[1]],['fouls',[4]],['formation_place',[2]]]],2,'DR',22,0,0,'D(CR)',30,173,64]

[83686,'Marc Bartra',8.31,[[['accurate_pass',[64]],['touches',[88]],['won_contest',[1]],['total_scoring_att',[1]],['aerial_won',[1]],['total_pass',[66]],['total_tackle',[5]],['aerial_lost',[1]],['fouls',[1]],['formation_place',[5]]]],2,'DC',15,0,0,'D(C)',22,181,70]

[13471,'Adriano',6.72,[[['accurate_pass',[16]],['touches',[28]],['aerial_won',[2]],['total_pass',[18]],['total_tackle',[1]],['formation_place',[3]]]],2,'DL',21,1,31,'D(CLR),M(LR)',29,172,67]

现在，我在上面的示例中手动完成的工作需要使用 PHP 可靠地完成。如您所见，每个玩家都有一组数据。为了将大字符串拆分为单个玩家，我不能只用 "],[" 分解它，因为该子字符串出现在每个玩家的数据中的次数也无法预测。

每个球员都有一定数量的统计数据（accurate_pass、touches 等），但他们的统计数据并不完全相同。例如，玩家 #1 有“保存”，而其他玩家则没有。玩家 #4 有“won_contest”，其他人没有。没有办法知道谁将拥有哪些统计数据。这意味着我不能只计算逗号，直到新玩家或类似的东西。

每个玩家的名字前都有一个数字，但该数字的位数无法预测，并且无法将其与字符串中可能出现的其他数字区分开来。

我认为所有玩家都会经常出现的是最后一位：在最后一个右括号之前，总是有 3 个整数除以逗号。这种类型的子字符串 ( INT,INT,INT]) 似乎不会出现在任何其他情况下。也许这可能会有一些用处？

4

5 回答 5

1

看起来@Boundless 答案是正确的，您可以使用 json_decode，但您需要对首先获得的字符串做几件事，这似乎也是一个有效的 json 格式字符串。

这对我有用：

<?php
$str = "[9484,'Víctor Valdés',8,[[['accurate_pass',[15]],['touches',[42]],['saves',[4]],['total_pass',[24]],['good_high_claim',[2]],['formation_place',[1]]]],1,'GK',1,0,0,'GK',31,183,78],[1320,'Carles Puyol',7.76,[[['accurate_pass',[50]],['touches',[75]],['aerial_won',[3]],['total_pass',[55]],['total_tackle',[1]],['formation_place',[6]]]],2,'DC',5,0,0,'D(CLR)',35,178,80],[5780,'Dani Alves',8.21,[[['accurate_pass',[58]],['touches',[99]],['total_scoring_att',[1]],['total_pass',[66]],['total_tackle',[6]],['aerial_lost',[1]],['fouls',[4]],['formation_place',[2]]]],2,'DR',22,0,0,'D(CR)',30,173,64],[83686,'Marc Bartra',8.31,[[['accurate_pass',[64]],['touches',[88]],['won_contest',[1]],['total_scoring_att',[1]],['aerial_won',[1]],['total_pass',[66]],['total_tackle',[5]],['aerial_lost',[1]],['fouls',[1]],['formation_place',[5]]]],2,'DC',15,0,0,'D(C)',22,181,70],[13471,'Adriano',6.72,[[['accurate_pass',[16]],['touches',[28]],['aerial_won',[2]],['total_pass',[18]],['total_tackle',[1]],['formation_place',[3]]]],2,'DL',21,1,31,'D(CLR),M(LR)',29,172,67]";
$str = '[' . $str . ']';
$str = str_replace('\'','"', $str);


//convert string to array
$arr = json_decode($str);

//now it's a php array so you can access any value
//echo '<pre>';
//print_r( $arr );
//echo '</pre>';

echo $arr [0][1]; //prints "Victor Valdes"
?>
于 2013-10-31T19:58:04.753 回答
1

一个“硬”的方法是括号计数（在 PHP 中不太常见，在文本解析语言中更常见）......

<?php
$str = "[9484,'Víctor Valdés',8,[[['accurate_pass',[15]],['touches',[42]],['saves',[4]],['total_pass',[24]],['good_high_claim',[2]],['formation_place',[1]]]],1,'GK',1,0,0,'GK',31,183,78],[1320,'Carles Puyol',7.76,[[['accurate_pass',[50]],['touches',[75]],['aerial_won',[3]],['total_pass',[55]],['total_tackle',[1]],['formation_place',[6]]]],2,'DC',5,0,0,'D(CLR)',35,178,80],[5780,'Dani Alves',8.21,[[['accurate_pass',[58]],['touches',[99]],['total_scoring_att',[1]],['total_pass',[66]],['total_tackle',[6]],['aerial_lost',[1]],['fouls',[4]],['formation_place',[2]]]],2,'DR',22,0,0,'D(CR)',30,173,64],[83686,'Marc Bartra',8.31,[[['accurate_pass',[64]],['touches',[88]],['won_contest',[1]],['total_scoring_att',[1]],['aerial_won',[1]],['total_pass',[66]],['total_tackle',[5]],['aerial_lost',[1]],['fouls',[1]],['formation_place',[5]]]],2,'DC',15,0,0,'D(C)',22,181,70],[13471,'Adriano',6.72,[[['accurate_pass',[16]],['touches',[28]],['aerial_won',[2]],['total_pass',[18]],['total_tackle',[1]],['formation_place',[3]]]],2,'DL',21,1,31,'D(CLR),M(LR)',29,172,67]";
$line = ',';
$paren_count = 0;
$lines = array();
for($i=0; $i<strlen($str); $i++)
{
    $line.= $str{$i};
    if($str{$i} == '[') $paren_count++;
    elseif($str{$i} == ']')
    {
        $paren_count--;
        if($paren_count == 0)
        {
            $lines[] = substr($line,1);
            $line = '';
        }
    }
}
print_r($lines);
?>
于 2013-10-31T19:44:58.490 回答
0

您的字符串看起来像 JSON，但它不是有效的 JSON，因此json_decode()不起作用。

通过将字符串包装成一对[]并将单引号替换为双引号，可以将您的特定情况转换为有效的 JSON：

$string = str_replace("'", '"', $your_string);
var_dump(json_decode('[' . $string . ']'));

请参阅此示例。

当然，最好的解决方案是确保提供有效的 JSON，因为如果您的文本字符串包含例如双引号，这将很容易破坏。

于 2013-10-31T19:47:57.633 回答
0

尝试解析为 json，然后取出你想要的。假设数据以 4 个为一组，您可以尝试：

$arr = json_decode($str);
for($i = 0; $i < count($arr) - 3; $i += 4)
{
  $arr[] = new array($arr[$i], $arr[$i + 1], $arr[$i + 2], $arr[$i + 3]);
}
于 2013-10-31T19:42:38.747 回答
0

为什么不[循环计算呢？这是一个未经测试的快速循环，可以帮助您入门。

$output = array('');
$brackets = 0;
$index = 0;
foreach (str_split($input) as $ch) {
    if ($ch == '[') {
        $brackets++;
    }

    $output[$index] .= $ch;

    if ($ch == ']') {
        $brackets--;
        if ($brackets === 0) {
            $index++;
            $output[$index] = '';
        }
    }
}

虽然不是很优雅...

于 2013-10-31T19:45:11.087 回答