我正在处理一个php脚本,它在其相应的用户帐户表中存储消息id(Msg_ID,Ref_ID)。选择一个auto_increment字段返回空白在php
我所知道的是,Msg_ID被正确写入,但Ref_ID始终为空。 如何当我分别运行查询它的作品,但不能在脚本中出于某种奇怪的原因。
下面是代码:
$qry = "SELECT Ref_ID FROM Chat WHERE Msg_ID = " .$MsgID. ")";
$resp = mysqli_query($con, $qry);
$xx = mysqli_fetch_array($resp);
$ref_id = $xx['Ref_ID'];
foreach ($Array as $user){
$query = "Insert into ".$user."(POST_ID, REF_ID) values ('". $MsgID . "', '" .$ref_id. "')";
mysqli_query($con, $query);
}
的$ REF_ID始终是空白,因此,空值被写入到相应的数据库。 一些有助于解决问题的方法将有所帮助。
下面是完整的代码:聊天表
<?php
function PostMainThread($Heading, $Message, $Author, $MarkedList){
$con=mysqli_connect("mysql.serversfree.com", "u521497173_root", "123456", "u521497123_mydb");
$Array = explode(',', $MarkedList);
if (mysqli_connect_errno()){
$response["success"] = 0;
$response["message"] = "Connection Failed.";
echo json_encode($response);
}else{
here:$MsgID = rand(1, 9999999);
$query = "Insert into Chat(Msg_ID, Header, MsgBody, Author) values (". $MsgID . "," . "'" . $Heading . "' ," .
"'" . $Message . "', '". $Author . "')";
$result=mysqli_query($con, $query);
if (!$result){
goto here;
}else{
//Put the MsgID in the respective user tables.
$qry = "SELECT Ref_ID FROM Chat WHERE Msg_ID = " .$MsgID. ")";
$resp = mysqli_query($con, $qry);
$xx = mysqli_fetch_array($resp);
$ref_id = $xx['Ref_ID'];
foreach ($Array as $user){
$query = "Insert into ".$user."(POST_ID, REF_ID) values ('". $MsgID . "', '" .$ref_id. "')";
mysqli_query($con, $query);
}
$response["success"] = 1;
$response["message"] = "Submission successful.";
mysqli_close($con);
echo json_encode($response);
}
}
}
function PostReplyToThread($PostID, $Author, $Reply){
$con=mysqli_connect("mysql.serversfree.com", "u521497123_root", "123456", "u521497123_mydb");
if (mysqli_connect_errno()){
echo 2;
}else{
$query = "Insert into Chat(Msg_ID, Header, MsgBody, Author) values (". $PostID . "," . "'" . " " . "' ," .
"'" . $Reply . "', '". $Author . "')";
$result=mysqli_query($con, $query);
if ($result){
echo 3;
}else{
echo 4;
}
mysqli_close($con);
}
}
if (isset($_POST['what_to_do'])){
if ($_POST['what_to_do'] == 0){
if ((isset($_POST['Title'])) &&(isset($_POST['Body']))&&(isset($_POST['Marked']))&&(isset($_POST['_Author']))){
PostMainThread($_POST['Title'], $_POST['Body'], $_POST['_Author'], $_POST['Marked']);
}
}else if ($_POST['what_to_do'] == 1){
if ((isset($_POST['Thread_ID'])) &&(isset($_POST['Answer']))&&(isset($_POST['_Author']))){
PostReplyToThread($_POST['Thread_ID'], $_POST['_Author'], $_POST['Answer']);
}
}
}else{
$response["success"] = 0;
$response["message"] = "Unspecified action";
echo json_encode($response);
}
定义:
Create table Chat(Ref_ID INT Auto_Increment, Msg_ID INT, Header varchar(50), MsgBody varchar(500
), Author varchar(30), Primary Key(Ref_ID, Msg_ID));
Em,你真的应该阅读关于错误处理......你是盲目的! – arkascha
另外,您应该阅读“准备好的陈述”的优点以及为什么他们的用法很重要。 – arkascha
嗯,我不明白这里出了什么问题。因为只有Ref_ID没有被写入数据库。 – Priyabrata