PHP PDO $ row內部php代碼未運行

Question

我有問題，我用$ row [name]在我的數據庫中存儲了一條消息。

 Helloo $row['name'] how are you 

$txt[1] is "Helloo $row['name'] how are you "

這是我嘗試的第一個回顯有效的代碼，但第二個不起作用

<?php
$check = $db->prepare("SELECT * FROM appointments WHERE trimis='0'"); 
$check->execute();
$checkdb = $check -> fetchAll();
foreach($checkdb as $row){

    echo $row['name']; // THIS WORKS

$check = $db->prepare("SELECT * FROM settings WHERE id='1'"); 
$check->execute();
$checkdb2 = $check -> fetchAll();
foreach($checkdb2 as $txt){
    echo $txt[1]."<BR>"; // THIS DOESEN'T
}

}
?>

Answer 1

編寫查詢時，應從表中選擇特定字段，因為使用通配符SELECT * +按索引訪問行（例如$txt[1]是解決錯誤的方法：

$check1 = $db->prepare("SELECT name FROM appointments WHERE trimis='0'"); 
$check1->execute();
while($rows1 = $check1->fetch()){
    echo $row['name'];
    $check2 = $db->prepare("SELECT custom_text FROM settings WHERE id='1'"); 
    $check2->execute();
    while($rows2 = $check2->fetch()){
        echo $rows2['custom_text']."<BR>"; 
    }
}