Question

因此，我有将这类项目分成几个省的项目。我有两个多维数组与table1和table2，我一直试图用foreach函数等回显它，但仍然出错。这是我的数组：

array(2) {
  [0]=>
  array(1) {
    [1]=>
    array(2) {
      ["table1"]=>
      array(12) {
        [0]=>
        string(1) "1"
        ["id_province"]=>
        string(1) "1"
        [1]=>
        string(13) "Province A"
        ["nm_province"]=>
        string(13) "Province A"
      }
      ["table2"]=>
      array(2) {
        [0]=>
        array(58) {
          [0]=>
          string(2) "43"
          ["id_news"]=>
          string(2) "43"
          [1]=>
          string(1) "1"
          ["id_province"]=>
          string(1) "1"
          [2]=>
          string(23) "News A"
          ["nm_news"]=>
          string(23) "News A"
        }
        [1]=>
        array(58) {
          [0]=>
          string(3) "123"
          ["id_news"]=>
          string(3) "123"
          [1]=>
          string(1) "1"
          ["id_province"]=>
          string(1) "1"
          [2]=>
          string(21) "News B"
          ["nm_news"]=>
          string(21) "News B"
        }
      }
    }
  }
  [1]=>
  array(1) {
    [2]=>
    array(2) {
      ["table1"]=>
      array(12) {
        [0]=>
        string(1) "2"
        ["id_province"]=>
        string(1) "2"
        [1]=>
        string(23) "Province B"
        ["nm_province"]=>
        string(23) "Province B"
      }
      ["table2"]=>
      array(2) {
        [0]=>
        array(58) {
          [0]=>
          string(2) "44"
          ["id_news"]=>
          string(2) "44"
          [2]=>
          string(1) "2"
          ["id_province"]=>
          string(1) "2"
          [5]=>
          string(24) "News A Province B"
          ["nm_news"]=>
          string(24) "News A Province B"
        }
        [1]=>
        array(58) {
          [0]=>
          string(3) "127"
          ["id_news"]=>
          string(3) "127"
          [2]=>
          string(1) "2"
          ["id_province"]=>
          string(1) "2"
          [5]=>
          string(13) "News B Province B"
          ["nm_news"]=>
          string(13) "News B Province B"
        }
      }
    }
  }
}

我不知道如何在数组中检索2个以上的表。所以我再说一遍，我想在第一张省有2条新闻和第二张省也有2条新闻回显这2个表，我要做的是在省份中回显这2条新闻。

Ps。这是我的代码，用于显示数组输出

<?php 
                        $a=mysql_query("select * from province");

                        while($m1=mysql_fetch_array($a)){
                            $result[]=$m1;
                        }
                        $output=[];
                        $i=0;
                        foreach($result as $r){
                            $b=$r['id_province'];
                            $c=mysql_query("select * from news where id_province=".$b);
                            $output[$i][$b]['table1']=$r;
                            $dummy=[];
                            while($response = mysql_fetch_array($c)){
                                $dummy[] = $response;
                            }
                            $output[$i][$b]['table2']=$dummy;
                            $i++;
                        }

感谢您的帮助。

Answer 1

更新以下代码，它将为您服务。

 foreach($result as $r){
            $b=$r['id_province'];
            $c=mysql_query("select * from news where id_province=".$b);
            $output[$i]['table1']=$r;
            $dummy=[];
            while($response = mysql_fetch_array($c)){
                $dummy[] = $response;
            }
            $output[$i]['table2']=$dummy;
            $i++;
        }

在这里，我从[$b]数组中删除了另外的$output。

现在复制下面的代码以打印数组。

foreach($output as $out){
            $table1= $out['table1'];
            $table2= $out['table2'];

            echo $table1['nm_provinsi'].'<br>';

            echo 'Data from table 2';
            foreach($table2 as $tab2){

                    echo "\t".$tab2['nm_berita'].'<br>';
            }

        }

Answer 2

我想说，您需要先清理数组，以确保尝试引用正确的对象，似乎所有内容都重复了，并带有索引和数组中的键。

[0]=>"123",
["id_news"]=>"123",

[1]=>"1",
["id_province"]=>"1",

[2]=>"News B",
["nm_news"]=>"News B",

// -----编辑----- //

我已经分解了数组，以便您可以清楚地理解我的意思

array(
    'table1' => array(
        '0' => 1,
        'id_province' => 1,

        '1' => 'Province A',
        'nm_province' => 'Province A'
    ),
    'table2' => array(
        array(
            '0' => '43',
            'id_news' => '43',

            '1' => '1',
            'id_province' => '1',

            '2' => 'News A',
            'nm_news' => 'News A'
        ),
        array(
            '0' => '123',
            'id_news' => '123',

            '1' => 1,
            'id_province' => '1',

            '2' => 'New B',
            'nm_news' => 'News B')
    )
),
array(
    'table1' => array(
        '0' => "2",
        'id_province' => "2",

        '1' => "Province B",
        'nm_province' => 'Province B'
    ),
    'table2' => array(
        array(
            '0'=>'44',
            'id_news' => '44',

            '2' => '2',
            'id_province' => '2',

            '5' => 'News A Province B',
            'nm_news' => 'News A Province B'
        ),
        array(
            '0'=>'127',
            'id_news' => '127',

            '2'=>'2',
            'id_province' => '2',

            '5' => 'News B Province B',
            'nm_news' => 'News B Province B'
        )
    )
)

Answer 3

尝试

    $c=mysql_query("select * from news where id_province='$b'");

回声两个具有不同表的多维数组

3 个答案: