如何在 php 数组中搜索 mysql 表中的所有项目？

Question

我只想显示 usernameArray 中不存在且仅存在于 mysql 表中的用户名。换句话说，对于 mysql 表中的每个用户名，我想在 usernamesArray 中检查它，如果该数组中不存在，我只是从 mysqltable 输出该用户名？谁能告诉我如何完成这项任务。谢谢

       $url = "https://api.instagram.com/v1/users/XXXX/follows?access_token=XXXX&count=-1";
       $api_response = get_data(''.$url);
       $record = json_decode($api_response); // JSON decode


$m = 0;

$usernamesArray = array();

foreach($record->data as $user) // each user data (JSON array) defined as $user
{
 $m++;

$usernameVar = $user->username;
$usernamesArray[] = $usernameVar;

}

print_r($usernamesArray);


$sql->Query("SELECT * FROM mytable");

echo "Total:".$sql->rows;
echo "<br>";
    for ($i = 0; $i < $sql->rows; $i++) 
    {
        $sql->Fetch($i);
        $id = $sql->data[0];
        $username = $sql->data[1];
        $website = $sql->data[2];
        $profile_picture = $sql->data[3];


//now compare usernamesArray with current data in mysql table and only display 
//the usernames that doesnt exist in usernamesArray?

 echo("<div id='grid-cell' style='padding:5px'><a style='text-decoration:none' href='$username'><img class='photo-grid' src='$profile_picture' width=150 height=150 title='$username' /></a><div class='moreInfo2'><a style='color:#000;text-decoration:none' href='/$username' target='_blank'>$item:$username()</a></div></div>\n");

   }

Answer 1

尝试改变：

echo("<div id='grid-cell' style='padding:5px'><a style='text-decoration:none' href='$username'><img class='photo-grid' src='$profile_picture' width=150 height=150 title='$username' /></a><div class='moreInfo2'><a style='color:#000;text-decoration:none' href='/$username' target='_blank'>$item:$username()</a></div></div>\n");

到：

if(!in_array($username, $usernamesArray))
{
echo("<div id='grid-cell' style='padding:5px'><a style='text-decoration:none' href='$username'><img class='photo-grid' src='$profile_picture' width=150 height=150 title='$username' /></a><div class='moreInfo2'><a style='color:#000;text-decoration:none' href='/$username' target='_blank'>$item:$username()</a></div></div>\n");
}

Answer 2

类似这样的事情（您需要先存储 SQL 行，然后检查 in_array）：

$usernamesArray = array();

$sql->Query("SELECT * FROM mytable");
for ($i = 0; $i < $sql->rows; $i++) {
    $sql->Fetch($i);
    $usernamesArray[] = $sql->data[1];
}

foreach ($record->data as $user) {
    $usernameVar = $user->username;
    if (!in_array($usernameVar,$usernamesArray)) {
        echo("<div id='grid-cell' style='padding:5px'><a style='text-decoration:none' href='$usernameVar'><img class='photo-grid' src='$profile_picture' width=150 height=150 title='$username' /></a><div class='moreInfo2'><a style='color:#000;text-decoration:none' href='/$usernameVar' target='_blank'>$item:$usernameVar()</a></div></div>\n");
    }
}