Question

我制作了一个电子邮件脚本，只要wp_mail有结果就应该更新。出于某种原因，我的价值不会更新。我错过了什么吗？我收到了邮件，所以wp_mail正常工作。

干杯！

$email_result = wp_mail( $to, $subject, $message, $headers );

if( $email_result ){//wp_mail() processed the request successfully
    global $wpdb;
    $table_name = $wpdb->prefix . "wpsc_coupon_codes";
    $coupon_id = $ereminder->ID;

$ereminders = $wpdb->query( $wpdb->prepare("
    UPDATE *
    FROM $table_name
    SET reminder = 1
    WHERE ID = $coupon_id
") );

}

Answer 1

试试这个：

$wpdb->update( $table_name, array( 'reminder' => 1),array('ID'=>$coupon_id));

Answer 2

试试这个

UPDATE  $table_name
SET reminer = 1
WHERE ID = $coupon_id

Answer 3

您可以更改而不是（UPDATE * FROM）

$ereminders = $wpdb->query($wpdb->prepare("UPDATE $table_name SET reminer='1' WHERE ID=$coupon_id"));

并且不用休息。

谢谢。

Answer 4

正在运作的矿井示例：

$result = $wpdb->update(
    $wpdb->prefix .'sae_calendar', 
    array( 
        'year' => $year,
        'quarter' => $quarter,
        'start_date' => $start_date,
        'end_date' => $end_date,
        'reservation_start_date' => $reservation_start_date,
        'reservation_end_date' => $reservation_end_date 
    ), 
    array(
        "id" => $id
    ) 
);

Answer 5

UPDATE $table_name
SET reminder = 1
WHERE ID = $coupon_id

另外，改变：

$coupon_id = $ereminder->ID;

到

$coupon_id = $ereminder->id;

来源：Update Query With $wpdb - Wordpress Tutorial

Answer 6

<?php 
  global $wpdb;
    if(isset($_POST['progress'])){
    $table=t_test;
    $data=array('client_development'=>$_POST['Progress']);
    $where=array('p_id'=>$_SESSION['id']);
    $format=("%d");
    $whereFormat=("%d");
    $result4=$wpdb->UPDATE($table,$data,$where,$format,$whereFormat);
  }
?>
<form method="post">
  <input type="text" name="Progress">
  <input type="submit" name="progress" value="Prog%">
</form>
<?php 
if(isset($_POST['progress'])){
$table=t_test;
    $data=array('client_development'=>$_POST['Progress']);
    $where=array('p_id'=>$_SESSION['id']);
    $format=("%d");
    $whereFormat=("%d");
    $result4=$wpdb->UPDATE($table,$data,$where,$format,$whereFormat);
}
?>
<form method="post">
  <input type="text" name="Progress">
  <input type="submit" name="progress" value="Prog%">
</form>

Answer 7

我们可以这样使用$wpdb->update：

 global $wpdb;
 $table_name = $wpdb->prefix.'your_table_name';
 $data_update = array('row_name_1' => $row_data_1 ,'row_name_2' => $row_data_2);
 $data_where = array('row_name' => $row_data);
 $wpdb->update($table_name , $data_update, $data_where);

wpdb更新查询不起作用

7 个答案: