Question

我正在尝试遍历从select * from table_name where id in (select min(id) from table_name group by c, s);结果获得的数组，以便从所有结果中创建一个字符串。

Array看起来像这样：

preg_match_all

我的代码：

print_r($matches[0]);


Array
(
    [0] => Array
        (
            [0] => 8147
            [1] => 3
        )

    [1] => Array
        (
            [0] => 8204
            [1] => 20
        )

)

我希望得到$found = count($matches[0]); for ($i = 0; $i <= $found; $i++) { $string = $matches[0][$i]; }的结果：$string

如何使用循环在字符串变量中将8147, 8204附加到$matches[0][0]等？

Answer 1

你可以这样做吗

$string = "";
foreach($matches[0] as $value) {
    $string .= $value[0].", ";
}
$string = rtrim(", ",$string);

Answer 2

尝试以下代码。循环遍历数组并获取值

$arr =Array
(
    0 => Array
        (
            0 => 8147,
            1 => 3
        ),

    1 => Array
        (
            0 => 8204,
            1 => 20
        )

);
$match_array = array();
foreach($arr as $key=>$value)
{
    $match_array[] = $value[0];
}
$str =  implode(",",$match_array);
echo $str;

DEMO

或者只是使用array_column将特定列作为数组获取然后使用implode

$arr =Array
(
    0 => Array
        (
            0 => 8147,
            1 => 3
        ),

    1 => Array
        (
            0 => 8204,
            1 => 20
        )

);
$match_array = array_column($arr,0);

$str =  implode(",",$match_array);
echo $str;

DEMO

Answer 3

使用php5.5及更多版本，您可以使用array_column + implode：

echo implode(', ', array_column($matches, 0));

Answer 4

您可以使用array_column，无需循环遍历数组

$result = join(',', array_column($arr, 0));

PHP循环遍历数组并将其元素转换为一个字符串

4 个答案: