下拉其他值不保存到数据库 [英] dropdown other value not saving to database
本文介绍了下拉其他值不保存到数据库的处理方法，对大家解决问题具有一定的参考价值，需要的朋友们下面随着小编来一起学习吧！
问题描述
< script type =text / javascript>
function showfield（name）{
if（name =='Other'）document.getElementById（'div1'）。innerHTML ='other：< input type =textname =other />';
else document.getElementById（'div1'）。innerHTML ='';
}
< / script>
< html>
< head>
< meta name =viewportcontent =width = device-width，initial-scale = 1.0>
< link href =css / bootstrap.min.css =stylesheet>
< title>插入< /标题>
< / head>
< body>
< h1>将数据插入到Db< / h1>
< form action =teachersubmit.phpmethod =postid =form>
< input type =hiddenname =submittedvalue =true/>
< label>学校编号：< input type =textname =scode/>< / label>
< label>分类：< input type =textname =category/>< / label>
< label>名称：< input type =textname =sname/>< / label>
< label>地址：< input type =textname =sadd>< / label>< br>
< label>名称：< select name =genderid =gender>
< option value =male> male< / option>
< option value =女性>女性< / option>
< option value =其他>其他< / option>
< / select>< / label>
< div id =div1>< / div>
< button type =submitform =formvalue =Submit> Submit< / button>
< / form>
if（isset（$ _ POST ['submitted']））{
$ con = mysqli_connect（localhost，root，，school）;
if（mysqli_connect_errno（））
{
echo无法连接到MySQL：。 mysqli_connect_error（）;
$ b if（！empty（$ _ POST ['other']））
{
$ sname = $ _ POST [ 'SNAME'];
$ scode = $ _ POST ['scode'];
$ category = $ _ POST ['category'];
$ sadd = $ _ POST ['sadd'];
$ gender = $ _ POST ['other'];
$ b $ sqlinsert =INSERT INTO form1（sname，scode，category，sadd，gender）VALUES（'$ sname'，'$ scode'，'$ category'，'$ sadd'，'$性别'）;
if（！mysqli_query（$ con，$ sqlinsert））{
die（'not inserted'）;
}}
else {
$ sname = $ _ POST ['sname'];
$ scode = $ _ POST ['scode'];
$ category = $ _ POST ['category'];
$ sadd = $ _ POST ['sadd'];
$ gender = $ _ POST ['gender'];
$ b $ sqlinsert =INSERT INTO form1（sname，scode，category，sadd，gender）VALUES（'$ sname'，'$ scode'，'$ category'，'$ sadd'，'$性别'）;
if（！mysqli_query（$ con，$ sqlinsert））{
die（'not inserted'）;
}
}
解决了.....谢谢所有你帮助我只是把这个条件，如果该隐藏领域是空的，然后采取下拉值，如果没有，然后采取文本框的值......干杯:)
这是数据库结构：
sname varchar（100）
category varchar（100）
sadd varchar （100）
scode varchar（20）
gender varchar（100）
选项中缺少值值，这是下拉值不张贴表单窗体的原因并且不保存到数据库中
This is the database structure:
< option> male< / option>
<选项>女性< / option>
这将是
<选项值=男性>男性< / option>
< option value =女性>女性< / option>
要检查查询中的错误，请阅读
http://php.net/manual/en/mysqli.error.php
您的代码已经打开，用于sql注入。阅读此内容以防止形成它
EDITED
数据库中没有插入数据，因为 sname，scode，category，sadd 是整数类型你试图插入字符
更改 sname，scode，category，sadd 作为 varchar 归档到您的数据库中
<script type="text/javascript">
function showfield(name){
if(name=='Other')document.getElementById('div1').innerHTML='other: <input type="text" name="other" />';
else document.getElementById('div1').innerHTML='';
}
</script>
<html>
<head>
<meta name="viewport" content="width=device-width, initial-scale=1.0">
<link href = "css/bootstrap.min.css" rel = "stylesheet">
<title> Insert</title>
</head>
<body>
<h1>Insert Data In to Db</h1>
<form action="teachersubmit.php" method="post" id="form">
<input type="hidden" name="submitted" value ="true" />
<label> School Code: <input type="text" name="scode" /></label>
<label> Category: <input type="text" name="category" /></label>
<label> name: <input type="text" name="sname" /></label>
<label> Address: <input type="text" name="sadd"></label><br>
<label> name: <select name="gender" id="gender">
<option value="male">male</option>
<option value="female">female</option>
<option value="Other">Other</option>
</select></label>
<div id="div1"></div>
<button type="submit" form="form" value="Submit">Submit</button>
</form>
if (isset($_POST ['submitted']) ){
$con = mysqli_connect("localhost","root","","school");
if (mysqli_connect_errno())
{
echo "Failed to connect to MySQL: " . mysqli_connect_error();
}
if(!empty($_POST['other']))
{
$sname=$_POST['sname'];
$scode=$_POST['scode'];
$category=$_POST['category'];
$sadd=$_POST['sadd'];
$gender=$_POST['other'];
$sqlinsert= "INSERT INTO form1 (sname,scode,category,sadd,gender) VALUES ('$sname' , '$scode', '$category', '$sadd', '$gender')";
if(!mysqli_query($con, $sqlinsert)){
die ('not inserted');
}}
else{
$sname=$_POST['sname'];
$scode=$_POST['scode'];
$category=$_POST['category'];
$sadd=$_POST['sadd'];
$gender=$_POST['gender'];
$sqlinsert= "INSERT INTO form1 (sname,scode,category,sadd,gender) VALUES ('$sname' , '$scode', '$category', '$sadd', '$gender')";
if(!mysqli_query($con, $sqlinsert)){
die ('not inserted');
}
}
Solved ..... thanks all of u for helping i just put the condition that if that hidden field is empty then take the dropdown value and if not then take the textbox value ......cheers :)
This is the database structure:
sname varchar(100)
category varchar(100)
sadd varchar(100)
scode varchar(20)
gender varchar(100)
解决方案
Value is missing in option that's the reason dropdown value not posting form form and not save into database
<option>male</option>
<option>female</option>
It would be
<option value="male">male</option>
<option value="female">female</option>
To check error in your query read
http://php.net/manual/en/mysqli.error.php
Your code is open for sql injection . Read this to prevent it form it
How can I prevent SQL injection in PHP?
EDITED
Data is not inserted because sname,scode,category,sadd are integer type in database and you are trying to insert character
ALter sname,scode,category,sadd filed as varchar into your database
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