php，html5，xampp
[英]Php, Html5, xampp
问题描述
I was trying execute this PHP file that I have in XAMPP. 
我试图执行XAMPP中具有的这个PHP文件。 I am trying to update the data to the database through a form, when I enter the info and give submit button the page just reloads or I get the message all fields are required though I fill up all. 我正在尝试通过表单将数据更新到数据库，当我输入信息并提供“提交”按钮时，页面将重新加载，或者我收到消息，尽管我填写了所有内容，但所有字段都是必需的。 Give an it a check to the code. 检查一下代码。 The data I enter just doesn't get updated to the database. 我输入的数据不会更新到数据库。 I've checked with the database name and table name that if I am mentioning it properly and also I've tried just checking that if I am able to connect to the database. 我已经检查了数据库名和表名，如果我正确地提到它，也尝试过仅检查是否能够连接到数据库。 In that case it is getting connected to the database properly.but just not updating the info the table. 在这种情况下，它可以正确连接到数据库。但是只是不更新表的信息。
<form action="" method="POST">
<div class="main">
<div class="icon-head-wthree">
<h2><span class="fa fa-heart t-w3" aria-hidden="true"></span></h2>
<h4>Donor?</h4>
</div>
<div class="form-left-to-w3l">
<input type="text" name="name" placeholder="First Name">
<div class="clear"></div>
</div>
<div class="form-left-to-w3l">
<input type="text" name="lname" placeholder="Last Name">
<div class="clear"></div>
</div>
<div class="form-left-to-w3l">
<input type="text" name="city" placeholder="City" >
<div class="clear"></div>
</div>
<div class="form-left-to-w3l">
<input type="text" name="district" placeholder="District" >
<div class="clear"></div>
</div>
<div class="form-left-to-w3l">
<input type="text" name="state" placeholder="State" >
<div class="clear"></div>
</div>
<div class="form-left-to-w3l">
<input type="text" name="Phone" placeholder="Phone 1" >
<div class="clear"></div>
</div>
<div class="form-left-to-w3l">
<input type="text" name="p1" placeholder="Phone 2">
<div class="clear"></div>
</div>
<div class="form-left-to-w3l">
<input type="text" name="pincode" placeholder="Pincode" >
<div class="clear"></div>
</div>
<div class="form-left-to-w3l">
<input type="text" name="bg " placeholder="Blood Group" >
<div class="clear"></div>
</div>
<div class="form-left-to-add">
<input type="text" name="add1 " placeholder="Address 1" >
<div class="clear"></div>
</div>
<div class="form-left-to-add">
<input type="text" name="add2 " placeholder="Address 2" >
<div class="clear"></div>
</div>
<input type="checkbox" name="check" checked="checked">
<span class="checkmark"></span>
<label class="cont">I agree to the terms and conditions
</label>
<div class="btnn">
<input type="submit" value="submit" name="submit">Submit Request </button><br>
<div class="clear"></div>
</div>
</div>
</form>
<?php
if(isset($_POST["submit"])){
$db= 'bank';
if(!empty($_POST['name']) && !empty($_POST['lname']) && !empty($_POST['city']) && !empty($_POST['district']) && !empty($_POST['state']) && !empty($_POST['Phone']) && !empty($_POST['p1']) && !empty($_POST['pincode']) && !empty($_POST['bg']) && !empty($_POST['add1']) && !empty($_POST['add2']))
{ //added
$name=$_POST['name'];
$lname=$_POST['lname'];
$city= $_POST['city'];
$district= $_POST['district'];
$state= $_POST['state'];
$Phone= $_POST['Phone'];
$p1= $_POST['p1'];
$pincode= $_POST['pincode'];
$bg= $_POST['bg'];
$add1= $_POST['add1'];
$add2= $_POST['add2'];
$con=mysqli_connect('localhost','root','',$db) or die(mysql_error());
$sql="INSERT INTO donor VALUES('$name','$lname','$city','$district','$state','$Phone','$p1','$pincode','$bg','$add1','$add2')";
$result=mysqli_query($con,$sql);
if($result){
echo "Account Successfully Created";
} else {
echo "Failure!";
}
}
else {
echo "All fields are required!";
}
}
?>
3 个解决方案
解决方案1
0 2018-03-18 04:51:06
Some of your input tag names has spaces like name="bg " instead of name="bg". 您的某些输入标签名称具有空格，例如name =“ bg”而不是name =“ bg”。 Same case for inputs - add1, and add2. 输入的大小写相同-add1和add2。
Updated code below. 更新了下面的代码。 It should be working fine. 它应该工作正常。
<form action="" method="POST">
<div class="main">
<div class="icon-head-wthree">
<h2><span class="fa fa-heart t-w3" aria-hidden="true"></span></h2>
<h4>Donor?</h4>
</div>
<div class="form-left-to-w3l">
<input type="text" name="name" placeholder="First Name">
<div class="clear"></div>
</div>
<div class="form-left-to-w3l">
<input type="text" name="lname" placeholder="Last Name">
<div class="clear"></div>
</div>
<div class="form-left-to-w3l">
<input type="text" name="city" placeholder="City" >
<div class="clear"></div>
</div>
<div class="form-left-to-w3l">
<input type="text" name="district" placeholder="District" >
<div class="clear"></div>
</div>
<div class="form-left-to-w3l">
<input type="text" name="state" placeholder="State" >
<div class="clear"></div>
</div>
<div class="form-left-to-w3l">
<input type="text" name="Phone" placeholder="Phone 1" >
<div class="clear"></div>
</div>
<div class="form-left-to-w3l">
<input type="text" name="p1" placeholder="Phone 2">
<div class="clear"></div>
</div>
<div class="form-left-to-w3l">
<input type="text" name="pincode" placeholder="Pincode" >
<div class="clear"></div>
</div>
<div class="form-left-to-w3l">
<input type="text" name="bg" placeholder="Blood Group" >
<div class="clear"></div>
</div>
<div class="form-left-to-add">
<input type="text" name="add1" placeholder="Address 1" >
<div class="clear"></div>
</div>
<div class="form-left-to-add">
<input type="text" name="add2" placeholder="Address 2" >
<div class="clear"></div>
</div>
<input type="checkbox" name="check" checked="checked">
<span class="checkmark"></span>
<label class="cont">I agree to the terms and conditions
</label>
<div class="btnn">
<input type="submit" value="submit" name="submit">Submit Request </button><br>
<div class="clear"></div>
</div>
</div>
</form>
<?php
if(isset($_POST["submit"])){
$db= 'bank';
if(!empty($_POST['name']) && !empty($_POST['lname']) && !empty($_POST['city']) && !empty($_POST['district']) && !empty($_POST['state']) && !empty($_POST['Phone']) && !empty($_POST['p1']) && !empty($_POST['pincode']) && !empty($_POST['bg']) && !empty($_POST['add1']) && !empty($_POST['add2']))
{ //added
$name=$_POST['name'];
$lname=$_POST['lname'];
$city= $_POST['city'];
$district= $_POST['district'];
$state= $_POST['state'];
$Phone= $_POST['Phone'];
$p1= $_POST['p1'];
$pincode= $_POST['pincode'];
$bg= $_POST['bg'];
$add1= $_POST['add1'];
$add2= $_POST['add2'];
$con=mysqli_connect('localhost','root','',$db) or die(mysql_error());
$sql="INSERT INTO donor VALUES('$name','$lname','$city','$district','$state','$Phone','$p1','$pincode','$bg','$add1','$add2')";
$result=mysqli_query($con,$sql);
if($result){
echo "Account Successfully Created";
} else {
echo "Failure!";
}
}
else {
echo "All fields are required!";
}
}
?>
解决方案2
0 2018-03-18 05:08:42
as the last answer remove the white space in (blood group : name="bg", address1: name="add1", address2: name="add2") some of name of input tag. 作为最后一个答案，请除去输入标记名称中的空白（血型：name =“ bg”，address1：name =“ add1”，address2：name =“ add2”）。
<form action="" method="POST">
<div class="main">
<div class="icon-head-wthree">
<h2><span class="fa fa-heart t-w3" aria-hidden="true"></span></h2>
<h4>Donor?</h4>
</div>
<div class="form-left-to-w3l">
<input type="text" name="name" placeholder="First Name">
<div class="clear"></div>
</div>
<div class="form-left-to-w3l">
<input type="text" name="lname" placeholder="Last Name">
<div class="clear"></div>
</div>
<div class="form-left-to-w3l">
<input type="text" name="city" placeholder="City" >
<div class="clear"></div>
</div>
<div class="form-left-to-w3l">
<input type="text" name="district" placeholder="District" >
<div class="clear"></div>
</div>
<div class="form-left-to-w3l">
<input type="text" name="state" placeholder="State" >
<div class="clear"></div>
</div>
<div class="form-left-to-w3l">
<input type="text" name="Phone" placeholder="Phone 1" >
<div class="clear"></div>
</div>
<div class="form-left-to-w3l">
<input type="text" name="p1" placeholder="Phone 2">
<div class="clear"></div>
</div>
<div class="form-left-to-w3l">
<input type="text" name="pincode" placeholder="Pincode" >
<div class="clear"></div>
</div>
<div class="form-left-to-w3l">
<input type="text" name="bg " placeholder="Blood Group" > // remoce white space
<div class="clear"></div>
</div>
<div class="form-left-to-add">
<input type="text" name="add1 " placeholder="Address 1" > // remoce white space
<div class="clear"></div>
</div>
<div class="form-left-to-add">
<input type="text" name="add2 " placeholder="Address 2" > // remoce white space
<div class="clear"></div>
</div>
<input type="checkbox" name="check" checked="checked">
<span class="checkmark"></span>
<label class="cont">I agree to the terms and conditions
</label>
<div class="btnn">
<input type="submit" value="submit" name="submit">Submit Request </button><br>
<div class="clear"></div>
</div>
</div>
</form>
<?php
if(isset($_POST["submit"]) == "submit"){
$db= 'bank';
if(!empty($_POST['name']) && !empty($_POST['lname']) && !empty($_POST['city']) && !empty($_POST['district']) && !empty($_POST['state']) && !empty($_POST['Phone']) && !empty($_POST['p1']) && !empty($_POST['pincode']) && !empty($_POST['bg']) && !empty($_POST['add1']) && !empty($_POST['add2']))
{ //added
$name=$_POST['name'];
$lname=$_POST['lname'];
$city= $_POST['city'];
$district= $_POST['district'];
$state= $_POST['state'];
$Phone= $_POST['Phone'];
$p1= $_POST['p1'];
$pincode= $_POST['pincode'];
$bg= $_POST['bg'];
$add1= $_POST['add1'];
$add2= $_POST['add2'];
$con=mysqli_connect('localhost','root','',$db) or die(mysql_error());
$sql="INSERT INTO donor VALUES('$name','$lname','$city','$district','$state','$Phone','$p1','$pincode','$bg','$add1','$add2')";
$result=mysqli_query($con,$sql);
if($result){
echo "Account Successfully Created";
} else {
echo "Failure!";
}
}
else {
echo "All fields are required!";
} }
?>
this code copy and paste. 此代码复制并粘贴。
and when ever you use the isset function please user the post's value to perfect condition like 当您使用isset函数时，请使用帖子的值以达到最佳状态，例如
if(isset($_POST["submit"]) == "submit") if（isset（$ _ POST [“ submit”]）==“ submit”）
if you have many button in one page then it will very help full thanks. 如果您在一页中有许多按钮，那么非常感谢。
解决方案3
0 2018-03-18 05:12:28
It is possible to write the INSERT INTO statement in two ways. 可以通过两种方式编写INSERT INTO语句。
The first way specifies both the column names and the values to be inserted: 第一种方法同时指定列名和要插入的值：
Step 1:
INSERT INTO table_name (column1, column2, column3, ...) VALUES (value1, value2, value3, ...);
If you are adding values for all the columns of the table, you do not need to specify the column names in the SQL query. 如果要为表的所有列添加值，则无需在SQL查询中指定列名。 However, make sure the order of the values is in the same order as the columns in the table. 但是，请确保值的顺序与表中各列的顺序相同。 The INSERT INTO syntax would be as follows: INSERT INTO语法如下：
Step 2:
INSERT INTO table_name VALUES (value1, value2, value3, ...);
And in your code I suppose you are trying to use the second method but you are not using all the column. 在您的代码中，我想您正在尝试使用第二种方法，但是您并未使用所有列。 (For example id column which may be auto increment ) （例如， id列可能是自动递增的）
So change your this insert code 因此，更改您的此插入代码
INSERT INTO donor VALUES('$name','$lname','$city','$district','$state','$Phone','$p1','$pincode','$bg','$add1','$add2');
to step 1 as I have shown above which looks like shown below. 转到我上面显示的步骤1 ，如下图所示。
INSERT INTO donor (name, lname, city, district, state, phone, p1, pincode, bg, add1, add2) VALUES('$name','$lname','$city','$district','$state','$Phone','$p1','$pincode','$bg','$add1','$add2');
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