提交表单时验证错误 [英] Validation error in submitting the form

问题描述

如果空名称提交表单，则表示显示错误消息。我使用一个html页面和一个php页面。我得到的是尽管名称不存在，但它将存储在其他数据中，如年龄到数据库中。我需要答案为文本字段一侧的[* Username missing]而不是下一个表格。 KIAND帮助请



我尝试了什么：



<？ php



$ link = mysqli_connect（localhost，root，，training）;



if（isset（$ _ POST ['submit']））

{



$ fname = $ _POST ['fname '];



if（空（$ _ POST [name]））{

$ fnameErr =Missing;

}

else {

$ fname =（$ _POST [fname]）;

}

$ sql =INSERT INTO record（Name）VALUES（'$ fname'）;



$ op = mysqli_query（$ link，$ sql ）;

if（$ op）{

echo学生信息已插入;

}

else {

echonot inserted;

}



}

mysqli_close（ $ link $;

？>

While submitting the form if empty name means error message to show. I using one html page and one php page . What i am getting is eventhough the name is not in also it will storing in other data given like age into database. I need the answer as [* Username missing ] on the side of textfield and not going to next form. KIndly help pls

What I have tried:

<?php

$link = mysqli_connect("localhost", "root", "", "training");

if(isset($_POST['submit']))
{

$fname = $_POST['fname'];

if (empty($_POST["name"])) {
$fnameErr = "Missing";
}
else {
$fname = ($_POST["fname"]);
}
$sql = "INSERT INTO record (Name) VALUES ('$fname')";

$op = mysqli_query($link,$sql);
if($op){
echo "Student info inserted";
}
else{
echo "not inserted";
}

}
mysqli_close($link);
?>

推荐答案

link = mysqli_connect（localhost，root，，training ）;



if（isset（

link = mysqli_connect("localhost", "root", "", "training");

if(isset(

_POST ['submit']））

{

_POST['submit']))
{

fname =

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