Question

我试图获取从数据库表中获取的数组中是否存在名称。该数组包含表members的全名字段但是，使用in_array()函数时，任何搜索结果始终为假

这是数组的样子

    Array ( [0] => Array ( [name] => Jane ) 
            [1] => Array ( [name] => Kelvin ) 
            [2] => Array ( [name] => James ) 
            [3] => Array ( [name] => Leon ) 
            [4] => Array ( [name] => Steve ) 
            [5] => Array ( [name] => Jenifer ) 
           )

这是我的 php代码，用于检查名称是否存在

<?php

    require_once 'databaseconn.php'; 
    #To get connection
    $mysqli = new mysqli($db_hostname, $db_username, $db_password,$db_database);
    #To get array of names from my database
    $stmt = $mysqli-> prepare('SELECT name FROM members WHERE id');
    $stmt->execute();
    $result = $stmt->get_result();
    #To store result as array in variable $application_names
    while ($row = $result->fetch_assoc()) {
        $member_names[] = $row;
    }
    #To test if name exists
    $name = "Jane";

    if (in_array($name, $member_names))
    {
    echo "Match found";
    }
    else
    {
    echo "Match not found";
    }

?>

Answer 1

<?php

    require_once 'databaseconn.php'; 
    #To get connection
    $mysqli = new mysqli($db_hostname, $db_username, $db_password,$db_database);
    #To get array of names from my database
    $stmt = $mysqli-> prepare('SELECT name FROM members WHERE id');
    $stmt->execute();
    $result = $stmt->get_result();
    #To store result as array in variable $application_names
    while ($row = $result->fetch_assoc()) {
        $application_names[] = $row;
    }
    #To test if name exists
    $name = "Jane";

    $flatArr = [];

    foreach ($member_names as $memberArr) {
        if(isset($memberArr['name'])) {
            $flatArr[] = $memberArr['name'];
        }
    }

    if (in_array($name,  $flatArr))
    {
    echo "Match found for: " . $name;
    }
    else
    {
    echo "Match not found for: " . $name;
    }

?>

PHP函数in_array（）未返回预期结果

1 个答案: