Php 如何插入多个ID';它在一个链接表中?
我想在Knowledgedetail中插入de EmployeeID和KnowledgeID。他创建了员工,但在知识细节方面什么也不做。我现在没有代码,我尝试了很多东西,但我不知道。Php 如何插入多个ID';它在一个链接表中?,php,jquery,mysql,Php,Jquery,Mysql,我想在Knowledgedetail中插入de EmployeeID和KnowledgeID。他创建了员工,但在知识细节方面什么也不做。我现在没有代码,我尝试了很多东西,但我不知道。 作为Addprofile.php中的第一个,您创建了概要文件,并且至少选择了您的知识。 我的问题是,如果一个人做了一个配置文件,并选择了如何在knowledgedetail中获取ID的知识 表1 员工:员工ID、姓名、机构、电子邮件、电话号码、照片、描述 表2 知识:知识ID,知识 表3 Knowledgedeta
作为Addprofile.php中的第一个,您创建了概要文件,并且至少选择了您的知识。 我的问题是,如果一个人做了一个配置文件,并选择了如何在knowledgedetail中获取ID的知识 表1
员工:员工ID、姓名、机构、电子邮件、电话号码、照片、描述
表2
知识:知识ID,知识
表3
Knowledgedetail:KnowledgedetailID、EmployeeID KnowledgeID EmployeeID out员工与EmployeeID out Knowledgedetail和
KnowledgeID out Knowledge与KnowledgeID out knowledgedetail有关系 Addprofile.php
<?php
include("connection.php");
?>
<!DOCTYPE html>
<html>
<head>
<title>Information System</title>
<link rel="stylesheet" type="text/css" href="css/test.css">
<meta charset ="utf-8">
<link rel='stylesheet' href='http://ajax.googleapis.com/ajax/libs/jqueryui/1.8/themes/smoothness/jquery-ui.css' type='text/css' media='screen' />
<link rel='stylesheet' href='css/ui.multiselect.css' type='text/css' media='screen' />
<script src="../Informatiesysteem/js/jquery.min.js"></script>
<script type='text/javascript' src='https://ajax.googleapis.com/ajax/libs/jquery/1.7.1/jquery.min.js'></script>
<script type='text/javascript' src='https://ajax.googleapis.com/ajax/libs/jqueryui/1.8.16/jquery-ui.min.js'></script>
<script type='text/javascript' src='../Informatiesysteem/js/ui.multiselect.js'></script>
<script type='text/javascript'>
jQuery(document).ready(function() {
jQuery("#selectitems").multiselect();
});
</script>
</head>
<body>
<div id="container">
<div id="logo"></div>
<div id="header">
<h1>Add Profile</h1>
</div>
<div id="menu">
</div>
<div id="content">
<?php
$result = mysql_query("select knowledgeid, knowledge from knowledge");
$items = array();
$selected = array();
while ($row = mysql_fetch_array($result)){
$id [] = $row [ 'knowlegdeid' ];
$items[] = $row[ 'knowledge' ];
}
//form processing
if (isset($_POST['selectitems'])) {
$selected = $_POST['selectitems'];
}
if (!empty($selected)) : print_r($selected); endif;
?>
<form enctype="multipart/form-data" id="my form" method="post" action="Addedprofile.php">
Name: <input type="text" name="name" /></br>
Establishment: <input type="text" name="establishment"/></br>
E-Mail: <input type="email" name="email"/></br>
Phonenumber: <input type="tel" name="phonenumber"/></br>
Photo: <input type="file" name="photo"/></br>
Description: <textarea rows="4" cols="50" name="description"></textarea></br>
Knowledge: <select name="selectitems[]" id="selectitems" multiple="multiple" style="width: 450px; height: 180px;">
<?php //first we add the list of selected items if any
foreach ($selected as $sel) { ?>
<option value="<?php echo $sel; ?>" selected="selected"><?php echo $id[$sel]; $items[$sel];?></option>
<?php } ?>
<?php foreach ($items as $i => $v) { //then insert all items, skipping those who were added above
if (in_array($d, $i, $selected)) : continue; endif; //skip ?>
<option value="<?php echo $d, $i; ?>"><?php echo $v; ?></option>
<?php } ?>
</select>
</br></br></br></br>
<input type="submit" name="add_profile" value="Add profile" />
</form>
</div>
</body>
</html>
<!DOCTYPE html>
<html>
<meta http-equiv="refresh" content=";URL=Addprofile.php" />
</html>
<?php
include ("connection.php");
$Name = $_POST['name'];
$Establishment = $_POST['establishment'];
$Email = $_POST['email'];
$Phonenumber = $_POST['phonenumber'];
$Photo = $_POST['photo'];
$Description = $_POST['description'];
$sql = "INSERT INTO employees
(
name,
establishment,
email,
phonenumber,
photo,
description
)
VALUES ('". $Name ."', '". $Establishment ."', '". $Email ."', '". $Phonenumber ."', '". $Photo ."', '". $Description ."')";
$sqldetail = "INSERT INTO knowledgedetail
(
employeeid,
knowledgeid
)
VALUES .......................";
$add = mysql_query($sql);
if ($add === false){
echo 'Profile is not created';
}
else {
echo "Profile created";
}
echo '</br>';
$knowledge = mysql_query($sqldetail);
if ($add === false){
echo 'Knowledge is not added';
}
else {
echo "Knowledge added";
}
echo '</br>';
?>
信息系统
jQuery(文档).ready(函数(){
jQuery(“#selectitems”).multiselect();
});
添加配置文件
姓名:
编制:
电子邮件:
电话号码:
照片:
说明:
知识:
您的代码有一个问题:
$knowledge = mysql_query($sqldetail);
if ($add === false){
echo 'Knowledge is not added';
}
else {
echo "Knowledge added";
}
在if
语句中,您应该比较$knowledge
而不是$add
。因此,它应该是:
$knowledge = mysql_query($sqldetail);
if ($knowledge === false){
echo 'Knowledge is not added';
}
else {
echo "Knowledge added";
}
另外,每次mysql\u query()
失败时,添加对mysql\u error()
的调用:
echo "MySQL ERROR: SQL = $sql -- Error=".mysql_error()";
您在第二次插入时忘记了值?我不知道应该是什么,我尝试了很多次,但都没有成功。您从未关闭$sqldetail,也从未执行$sqldetail。。。除非你没有给我们所有的密码我知道但不知道会发生什么