在Mysql数据库和显示数据提交的消息中插入动态选择框值

时间:2017-05-22 18:50:52

标签: php jquery mysql ajax dynamic

在下面的表格中,我根据资产标题值从mysql数据库加载问题和答案,并向数据库提交动态自定义问题值。 (这很好) 但我的问题是,当表单值被提交到数据库时,如果资产的自定义问题少于15个,我就不会收到消息。

我想我在 createlead.php 文件中做错了,如果条件......

请帮忙。

如果您认为可以通过其他方法解决此问题,请告知我们。

enter image description here enter image description here   根据资产标题值从mysql数据库加载动态自定义问题

<?php 

require_once 'db_config.php';

//if form is submitted
if($_POST) {    
$validator = array('success' => false, 'messages' => array());

$assettitle= $_POST['assettitle'];
$customquestion1= $_POST['cq1'];
$customquestion2= $_POST['cq2'];
$customquestion3= $_POST['cq3'];
$customquestion4= $_POST['cq4'];
$customquestion5= $_POST['cq5'];
$customquestion6= $_POST['cq6'];
$customquestion7= $_POST['cq7'];
$customquestion8= $_POST['cq8'];
$customquestion9= $_POST['cq9'];
$customquestion10= $_POST['cq10'];
$customquestion11= $_POST['cq11'];
$customquestion12= $_POST['cq12'];
$customquestion13= $_POST['cq13'];
$customquestion14= $_POST['cq14'];
$customquestion15= $_POST['cq15'];





$sql = "INSERT INTO ltfs (assettitle, customquestion1, customquestion2, customquestion3, customquestion4, customquestion5, customquestion6, customquestion7, customquestion8, customquestion9, customquestion10, customquestion11, customquestion12, customquestion13, customquestion14, customquestion15) VALUES ('$assettitle', '$customquestion1', '$customquestion2', '$customquestion3', '$customquestion4', '$customquestion5', '$customquestion6', '$customquestion7', '$customquestion8', '$customquestion9', '$customquestion10', '$customquestion11', '$customquestion12', '$customquestion13', '$customquestion14', '$customquestion15')";


$query = $connect->query($sql);


if($query === TRUE) {
    $validator['success'] = true;
    $validator['messages'] = "Your lead successfully submited to quality team";     
} else {        
    $validator['success'] = false;
    $validator['messages'] = "Error while adding the lead information";
}

// close the database connection
$connect->close();

echo json_encode($validator);

}

Jquery(将表单字段值发送到php文件): -

name = []
for line in lines: #get each element out of list
    name.append(line.loc_name) #get name of the line
    value=line.GetAttribute('c:loading') 
    print('Loading of the line: %s = %.2f%%'%(name,value))

createlead.php(将表单字段值提交到数据库): -

{{1}}

1 个答案:

答案 0 :(得分:0)

我认为这里犯了错误

if($query === TRUE)

你可以这样做

if($query){//you message}else{//error message}