Php in_array() 未按预期工作

Question

我目前正在处理多个 select 下拉列表。我需要在编辑表单上显示 selected 值。

in_array() 没有像我预期的那样工作，我的逻辑有错误吗？

显示数据库中 selected 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 通过以下方式获取：

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 通过以下方式获得：

$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();
    }
}

请帮我指明正确的方向。

我编辑了 get_artistsID_for_video 如下：

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;

这会将您的 ID 与条件相匹配。现在看起来你的 $current_artist_list 是具有键的关联数组值。尝试按我上面提到的那样仅推送值，或者如下更改代码。

$artists = array();
        if ($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 — Checks if a value exists in an array

那么，是什么检查了您的 in_array($value['id_artist'], $current_artist_list, true)？

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

但值 20 在您的 $current_artist_list 数组中 NOT。

因为 $current_artist_list 中每个元素的格式都是 array('id_artist_fk' => $row['id_artist_fk'])。

因此，您正在搜索 20，但您存储的值是 ('id_artist_fk' => 20).

20 不等于 array.

修复在 get_artistsID_for_video():

while ($row = $result->fetch_array()) {
    $artists[] = $row['id_artist_fk'];
}

现在您在数组中搜索 20，其中每个元素也是一个数字。

使您的搜索更快（仍在get_artistsID_for_video）：

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

并将in_array替换为：

// check for existence of a key, not value.
if (!empty($current_artist_list[$value['id_artist']])) {
    $selected = "selected='selected'";
}

Php in_array() 未按预期工作

Php in_array() not working as expected

html

php

foreach

multiple-select