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我正在使用收件箱系統。在前端,它使用jQuery和Ajax,因此不會刷新頁面。我已經處理了這個部分。在後端,有3個表格(現在)可以插入數據。PHP準備好的語句將數據插入到除一個表以外的所有表中(MySQL)
這裏是關係結構的基本概要:
conversations:
conversation_id int(11) primary key
conversation_subject varchar(128)
conversations_members:
conversation_id int(11)
user_id int(11)
conversation_last_view int(10)
conversation_deleted int(1)
conversations_messages:
message_id int(11) primary key
conversation_id int(11)
user_id int(11)
message_date timestamp
message_text text
因爲SENDER_ID始終爲0有一個額外的問題,但必須是另一個問題,因爲這是題外話。 問題在於conversations_members表。其他所有內容都會輸入到對話和conversations_messages表格中。這裏是PHP。問題是在底部的最後一個SQL查詢:
<?php
ini_set('display_errors', 1);
ini_set('display_startup_errors', 1);
error_reporting(E_ALL);
include('../inc/connect.php');
if (!isset($_SESSION['username'])) {
session_start();
}
$recipient_username = "";
$sender_id = "";
$a = 0;
$b = 0;
if(isset($_POST['subject'], $_POST['msg_body']) && !empty($_POST['subject']) && !empty($_POST['msg_body'])) {
//get ID of sender
$sender_id_query = "SELECT id FROM `users` WHERE username = ?";
$stmt = $connection->prepare($sender_id_query);
$stmt->bind_param('s', $_SESSION['username']);
$stmt->execute();
$result = mysqli_query($connection, $sender_id_query);
if($result) {
while($row = mysqli_fetch_assoc($result)) {
//$row['id'] = $sender_id; //neither of these work
$sender_id = $connection->insert_id; //Always zero
}
}
$stmt->close();
//get username of recipient
$recipient_name_query = "SELECT * FROM `users`";
$result = mysqli_query($connection, $recipient_name_query);
if($result) {
while($row = mysqli_fetch_assoc($result)) {
$row['username'] = $recipient_username;
}
}
//define post variables
$msg_subject = $_POST['subject'];
$msg_body = $_POST['msg_body'];
$subject = $connection->real_escape_string(htmlentities($msg_subject));
$body = $connection->real_escape_string(htmlentities($msg_body));
$conversation_id = mysqli_insert_id($connection);
//GET RECIPIENT ID
$sql = "SELECT id FROM `users` WHERE username=?";
$stmt = $connection->prepare($sql);
$stmt->bind_param('s', $recipient_username);
$result = mysqli_query($connection, $sql);
if ($result) {
while ($row = mysqli_fetch_assoc($result)) {
$recipient_id = $row['id'];
}
}
$stmt->close();
//INSERT SUBJECT INTO CONVERSATIONS TABLE
$stmt = $connection->prepare("INSERT INTO `conversations` (conversation_subject) VALUES(?)");
$stmt->bind_param('s', $subject);
$stmt->execute();
$stmt->close();
//INSERT THE IDs AND TIMESTAMPS INTO MESSAGES TABLE
$stmt = $connection->prepare("INSERT INTO `conversations_messages` (conversation_id, user_id, message_date, message_text)
VALUES(?, ?, NOW(), ?)");
$stmt->bind_param('iis', $conversation_id, $sender_id, $body);
$stmt->execute();
$stmt->close();
/*
THE FOLLOWING DATA DOES NOT GET INSERTED.....
*/
//INSERT IDs, LAST_VIEWED, AND DELETED INTO MEMBERS TABLE
$stmt = $connection->prepare("INSERT INTO `conversations_members` (conversation_id, user_id, conversation_last_view, conversation_deleted)
VALUES (?, ?, ?, ?)");
$stmt->bind_param('iiii', $conversation_id, $recipient_id, $a, $b);
$stmt->execute();
$stmt->close();
}
我沒有得到任何錯誤,我沒有看到任何錯別字。我哪裏做錯了?
嘗試調試: '$ Q = $ stmt-> bind_param( 'IIII',$ conversation_id,$ recipient_id,$一個,$ b)中;'' 的var_dump($ Q); var_dump($ stmt-> error)' –
添加'ini_set('display_errors',1);函數ini_set( 'log_errors',1);使用error_reporting(E_ALL); mysqli_report(MYSQLI_REPORT_ERROR | MYSQLI_REPORT_STRICT);'到腳本的頂部。這將強制任何'mysqli_'錯誤到 生成一個你不能錯過或忽略的異常。 – RiggsFolly
'mysqli_fetch_assoc()'返回的數組不是對象,所以'$ sender_id = $ row ['id']' – RiggsFolly