Question

我目前正在处理多选下拉列表。我需要在编辑表单上显示所选的值。

in_array()没有按照我的预期工作，我的逻辑中是否有错误？

显示数据库中所选值的代码是：

<div class="form-group">
    <label class="col-md-3 control-label" for="selectbasic">Update Artist Selection</label>
    <div class="col-md-6">
        <select id="artists1" multiple="multiple" name="id_artist_fk[]" class="form-control ">
            <?php
            foreach ($artist_list as $key => $value) {
                if (in_array($value['id_artist'], $current_artist_list, true)) {
                    $selected = "selected='selected'";
                }
                // print_r($value['id_artist']. "==". $current_artist_list);
                echo "<option value=\"{$value['id_artist']}\" {$selected}>{$value['name']} {$value['surname']}</option>";
            }
            ?>
        </select>
    </div>
</div>

$artist_list来自：

public function get_artist_list() {
    $sql = "SELECT * FROM tbl_v_artist WHERE status != 0;";
    $result = $this->database->doSelectQuery($sql);
    $artists = array();
    if ($result->num_rows > 0) {
        while ($row = $result->fetch_array()) {
            $artist = array(
                'id_artist' => $row['id_artist'],
                'name' => $row['name'],
                'surname' => $row['surname'],
                'status' => $row['status']
            );
            array_push($artists, $artist);
        }
    }
    return $artists;
}

$current_artist_list来自：

$current_artist_list = $vid->get_artistsID_for_video($_POST['id_video']);

get_artistsID_for_video是：

public function get_artistsID_for_video($video_id) {
    try {
        $sql = "SELECT
                tbl_video_artist.id_artist_fk
                FROM tbl_video_artist
                left join tbl_v_artist
                ON tbl_v_artist.id_artist = tbl_video_artist.id_artists
                WHERE tbl_video_artist.id_video_fk = {$video_id};";
        //echo $sql;
        $result = $this->database->doSelectQuery($sql);
        $artists = array();
        if ($result->num_rows > 0) {
            while ($row = $result->fetch_array()) {
                $artist = array(
                    'id_artist_fk' => $row['id_artist_fk']
                );
                array_push($artists, $artist);
            }
        }
        return $artists;
    } catch (Exception $ex) {
        $ex->getMessage();
        $ex->getFile();
    }
}

请指点我正确的方向。

我已按如下方式编辑了get_artistsID_for_video：

public function get_artistsID_for_video($video_id) {
    try {
        $sql = "SELECT
                tbl_video_artist.id_artist_fk
                FROM tbl_video_artist
                left join tbl_v_artist
                ON tbl_v_artist.id_artist = tbl_video_artist.id_artists
                WHERE tbl_video_artist.id_video_fk = {$video_id};";
        //echo $sql;
        $result = $this->database->doSelectQuery($sql);
       // $artists = array();
        if ($result->num_rows > 0) {
            while ($row = $result->fetch_array()) {
                $artist [] = $row['id_artist_fk'];
               // array_push($artists, $artist);
            }
             return $artist;
        }

    } catch (Exception $ex) {
        $ex->getMessage();
        $ex->getFile();
    }
}

Answer 1

请检查$ current_artist_list必须是单个数组。即$ current_artist_list [0] = 1; $ current_artist_list [1] = 2;

哪个会将您的ID与条件匹配。现在看起来你的$ current_artist_list是带有键的关联数组值。尝试按上面提到的仅推送值或更改下面的代码。

$artists = array();
        if ($result->num_rows > 0) {
            while ($row = $result->fetch_array()) {
                $artist = array(
                    'id_artist_fk' => $row['id_artist_fk']
                );
                array_push($artists, $row['id_artist_fk']);
            }
        }

Answer 2

查看in_array手册：

in_array - 检查数组中是否存在值

那么，检查你的in_array($value['id_artist'], $current_artist_list, true)是什么？

它检查$current_artist_list中是否存在$value['id_artist']的值。例如，如果$value['id_artist'] 20 ，in_array会检查数据 20 是否在您的数组中。

但 20 中的值 <{1}}数组中。

因为$current_artist_list中每个元素的格式为$current_artist_list。

因此，您正在搜索 20 ，但您存储的值为array('id_artist_fk' => $row['id_artist_fk'])。

('id_artist_fk' => 20) NOT EQUALS 20。

此修复程序位于array：

get_artistsID_for_video()

现在您在数组中搜索while ($row = $result->fetch_array()) { $artists[] = $row['id_artist_fk']; }，其中每个元素都是数字。

让您的搜索更加更快（仍在20中）：

get_artistsID_for_video

将while ($row = $result->fetch_array()) { // create array key with value of artist id $artists[$row['id_artist_fk']] = 1; }替换为：

in_array

Php in_array（）没有按预期工作

2 个答案: