PHP代码没有错误执行，插入操作成功，但数据

PHP代码下面是插入表格的详细信息到我的MySQL数据库，但不知道是哪里的问题不存在。显示没有错误，并显示一条消息“插入操作成功”PHP代码没有错误执行，插入操作成功，但数据

PHP代码下面是将表单详细信息插入到我的MySQL数据库中，但是不知道问题出在哪里。没有错误的显示和消息显示“插入操作成功”

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<html> 
<head> 
<meta charset="utf-8"> 
<title>Nihanth-HomePage</title> 
<link rel="shortcut icon" href="favicon.ico" type="image/x-icon"/> 
<link rel="stylesheet" href="css/bootstrap.min.css" /> 
<style> 
#body{ 
    background-image:url(images/diamond.jpg); 
} 
#navbar{ 
    margin-bottom:0px; 
    background-color:black; 
    color:white; 
} 
#aaa{ 
    background:#FFFFFF; 
} 
</style> 
</head> 
<body id="body"> 
<nav id="navbar" class="navbar navbar-default"> 
<div class="container-fluid"> 
<div class="navbar-header"> 
<a class="navbar-brand" href="/">Nihanth</a> 
</div> 
<ul class="nav navbar-nav navbar-right"> 
<li><a href="/">Home</a></li> 
<li><a href="about.php">About Me</a></li> 
<li><a href="myprojects.php">My Projects</a></li> 
<li><a href="contact.php">Contact Me</a></li> 
</ul> 
</div> 
</nav> 
<div class="container-fluid" id="aaa"> 
<?php 
$server_host="localhost"; 
$server_user="*******"; 
$db_name="********"; 
$server_pass="******"; 
$table="Medha_2k16"; 
$con = mysqli_connect($server_host, $server_user, $server_pass, $db_name) or die("Server Connection Failed"); 
// Check connection 
if (mysqli_connect_errno()) 
    { 
    echo "Failed to connect to MySQL: " . mysqli_connect_error(); 
    } 
?> 
<?php 
$sname=$_POST['sname']; 
$pin=$_POST['pin']; 
$phone=$_POST['phone']; 
$email=$_POST['email']; 
$branch=$_POST['branch']; 
$college=$_POST['college']; 
$quiz=$_POST['quiz']; 
$cd=$_POST['cd']; 
$md=$_POST['md']; 
$iwd=$_POST['iwd']; 
$elocution=$_POST['elocution']; 
$rp=$_POST['rp']; 
$ep=$_POST['ep']; 
$sp=$_POST['sp']; 
$pp=$_POST['pp']; 
$ppt=$_POST['ppt']; 
echo $sname;echo "<br><br>"; 
echo $pin;echo "<br><br>"; 
echo $phone;echo "<br><br>"; 
echo $email;echo "<br><br>"; 
echo $branch;echo "<br><br>"; 
echo $college;echo "<br><br>"; 
echo $quiz;echo "<br><br>"; 
echo $cd;echo "<br><br>"; 
echo $iwd;echo "<br><br>"; 
echo $elocution;echo "<br><br>"; 
echo $rp;echo "<br><br>"; 
echo $ep;echo "<br><br>"; 
echo $sp;echo "<br><br>"; 
echo $pp;echo "<br><br>"; 
echo $ppt;echo "<br><br>"; 

mysqli_query($con , "INSERT INTO Medha_2k16 (sname, pin, email, phone, branch, college, quiz, cd, iwd, elocution, sp, md, rp, ep, ppt, pp) VALUES ($sname,$pin,$email,$phone,$branch,$college,$quiz,$cd,$iwd,$elocution,$sp,$md,$rp,$ep,$ppt,$pp)"); 

if((mysqli_query)==true) 
{ 
    printf("Insert Operation Successful");  
} 
else 
{ 
    printf(" Unable to INSERT\n %d ",mysqli_error($con)); 
} 
mysqli_error($con); 
mysqli_close($con); 
?> 
</div> 
</body> 
</html>

来源

2016-08-14 Nihanth Charan

请您检查出的问题存在，你已经发布的代码没有表单元素可言。通过编辑问题将表单与此页面一起发布 –

您所使用的查询字符串里面 php variables

，因此它被视为一个string，不variable。他们的价值观无法使用。相反，您需要将其与查询字符串分开，以便可以处理它。

替换：

mysqli_query($con , "INSERT INTO Medha_2k16 (sname, pin, email, phone, branch, college, quiz, cd, iwd, elocution, sp, md, rp, ep, ppt, pp) VALUES ($sname,$pin,$email,$phone,$branch,$college,$quiz,$cd,$iwd,$elocution,$sp,$md,$rp,$ep,$ppt,$pp)");

有：

mysqli_query($con , "INSERT INTO Medha_2k16 (sname, pin, email, phone, branch, college, quiz, cd, iwd, elocution, sp, md, rp, ep, ppt, pp) VALUES (" '.$sname.'"," '.$pin.'"," '.$email.'"," '.$phone.'"," '.$branch.'"," '.$college.'"," '.$quiz.'"," '.$cd.'"," '.$iwd.'"," '.$elocution.'"," '.$sp.'"," '.$md.'"," '.$rp.'"," '.$ep.'"," '.$ppt.'"," '.$pp.'")");

这应该为你工作。

来源

2016-08-14 18:39:54

PHP代码没有错误执行，插入操作成功，但数据

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