我想将 Evenement 表的 id 发布到 Evenementontvanger.idEvent 列中。我尝试使用 mysql_insert_id() 但这不起作用。我尝试过这样的:
<?php
//include db configuration file
include 'connection.php';
function user_joined($user_werknemer,$user_project,$user_klant,$user_taak,$user_name,$user_desc, $user_start, $user_startdate, $user_starttime, $user_end, $user_enddate, $user_endtime, $user_color){
$q = "INSERT INTO evenement (id,title,description,start,startdate,starttime,end,enddate,endtime,color) VALUES
('','".$user_name."','".$user_desc."','".$user_start."','".$user_startdate."','".$user_starttime."','".$user_end."','".$user_enddate."','".$user_endtime."','".$user_color."') ";
$qo = "INSERT INTO evenementontvanger (idWerknemer,idProject,idEvent,idKlant,idTaak) VALUES ('".$user_werknemer."','".$user_project."','','".$user_klant."','".$user_taak."')";
mysql_query($q);
$id = mysql_insert_id();
$query = 'INSERT INTO evenementontvanger("idEvent") VALUES('.$id.')';
mysql_query($qo);}
if(isset($_POST['user_werknemer'],$_POST['user_project'],$_POST['user_klant'],$_POST['user_taak'],$_POST['user_name'],$_POST['user_desc'],$_POST['user_startdate'],$_POST['user_start'],$_POST['user_starttime'],$_POST['user_enddate'],$_POST['user_endtime'],$_POST['user_end'],$_POST['user_color'],$_POST['action'])){
$user_werknemer=$_POST['user_werknemer'];
$user_color=$_POST['user_color'];
$user_name=$_POST['user_name'];
$user_desc=$_POST['user_desc'];
$user_project=$_POST['user_project'];
$user_klant=$_POST['user_klant'];
$user_taak=$_POST['user_taak'];
$user_start=$_POST['user_startdate']." ".$_POST['user_starttime'];
$user_startdate=$_POST['user_startdate'];
$user_starttime=$_POST['user_starttime'];
$user_end=$_POST['user_enddate']." ".$_POST['user_endtime'];
$user_enddate=$_POST['user_enddate'];
$user_endtime=$_POST['user_endtime'];
$action=$_POST['action'];
if ($action=='joined'){
user_joined( $user_werknemer, $user_project, $user_klant, $user_taak, $user_name, $user_desc, $user_start, $user_startdate, $user_starttime, $user_end, $user_enddate, $user_endtime, $user_color);
}
}
/*if ( (isset($_POST["id"]) && strlen($_POST["id"]) >= 3 && strlen($_POST["id"]) <= 60) &&
(isset($_POST["name"]) && strlen($_POST["name"]) >= 3 && strlen($_POST["name"]) <= 50) &&
(isset($_POST["age"]) && strlen($_POST["age"]) >= 3 && strlen($_POST["age"]) <= 40) )
{ //check $_POST["name"] and $_POST["address"] and $_POST["city"] are not empty
$id = $_POST["id"];
$name = $_POST["name"];
$age = $_POST["age"];
$q = "INSERT INTO tbltest ( id, name, age) VALUES
('".$id."','".$name."','".$age."')";
mysql_query($q);
}*/
?>
我还尝试使用last_insert_id。但这并没有将 id 放入 idEvent 中。
最佳答案
在以下查询中,
$qo = "INSERT INTO evenementontvanger
(idWerknemer,idProject,idEvent,idKlant,idTaak)
VALUES
('".$user_werknemer."', '".$user_project."',
'', '".$user_klant."', '".$user_taak."')";
您正在从字段 idEvent
设置空字符串 ''
。
您可以直接使用 LAST_INSERT_ID()
来代替它,它是在执行为 $q
定义的查询后生成的;
喜欢:
$qo = "INSERT INTO evenementontvanger
(idWerknemer,idProject,idEvent,idKlant,idTaak)
VALUES
('".$user_werknemer."', '".$user_project."',
LAST_INSERT_ID(), '".$user_klant."', '".$user_taak."')";
执行mysql_query($q);
后,
以下陈述不是必需的。
$id = mysql_insert_id();
$query = 'INSERT INTO evenementontvanger("idEvent") VALUES('.$id.')';
关于php - 我想在发布后获得一个 id(自动增量)并将其发布到其他表中,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/22474133/