Querying data into phpmyadmin database table

时间:2016-04-04 18:35:38

标签: php mysql

I am trying to input the information from the form html into mysql table. I keep getting an error code but I do not understand it and I tried to research for solution but i came up short.

I put security because protecting my login information.

Please help me?

Here is the html:

<!DOCTYPE HTML>
<html>
<head>
<title> form</title>

<style type="text/css/css">
h2{
    text-align: center;
    margin-top: 2cm;
}

</style>
</head>

<body>
<form action='formDB2.php' method='POST'>

<p> Your Last Name: <input type="text" name="lastname" id="lastname" value="" size="30" /></p>

<p> Your First Name: <input type="text" name="firstname" id="firstname" value="" size="30" /></p>

      <p>Age: <input name="age" type="text " id="age "/>  </p>

     <p>gender: <select name="sex" id="gender">
     <option> Male </option>
     <option>Female </option>

     </select> </p>

<hr />

<p><input type="reset" value="REST"/><input type="submit" name="Submit" value="Submit" /></p>



</form>


</body>
</html>

Here the mysql/php code:

<?php

$link = mysqli_connect("localhost","*security*","*security*","*security*");

if($link === false){
    die("ERROR: Could not connect. " . mysqli_connect_error());
}



$lastname = mysqli_real_escape_string($link, $_POST['lastname']);
$firstname = mysqli_real_escape_string($link, $_POST['firstname']);
$age =  mysqli_real_escape_string($link, $_POST['age']);
$gender =  mysqli_real_escape_string($link, $_POST['sex']);



$sql = "INSERT INTO trainer(trainer_id, lastname, firstname, age, gender) VALUES (0,'$firstname', '$lastname', '$age' , '$gender')";
if(mysqli_query($link,$sql)){
   echo "Records added successfully.";
} else{
    echo "ERROR: Could not able to execute $sql. " . mysqli_error($link);
}





mysqli_close($link);
?>

Here is the output:

enter image description here

Here the account I want it to display:

enter image description here

1 个答案:

答案 0 :(得分:0)

trainer_id是您的主键,因此它必须是唯一的。您的代码始终设置为0:

$sql = "INSERT INTO trainer(trainer_id, lastname, firstname, age, gender) VALUES (0,'$firstname', '$lastname', '$age' , '$gender')";

您应该使用自动增量http://dev.mysql.com/doc/refman/5.7/en/example-auto-increment.html设置字段,并且不要在代码上传递任何内容:

$sql = "INSERT INTO trainer(lastname, firstname, age, gender) VALUES ('$firstname', '$lastname', '$age' , '$gender')";