Question

PHP代码下面是将表单详细信息插入我的MySQL数据库，但不知道问题出在哪里。未显示错误，并显示消息“插入操作成功”

<!doctype html>
<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>

Answer 1

您在查询字符串中使用了php variables，因此将其视为string，而不是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，以便可以对其进行处理。

<强>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)");

<强>与

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.'")");

这应该适合你。

PHP代码没有执行错误，插入操作成功，但

1 个答案: