使用 PHP 将用户添加到数据库时遇到问题
我正在尝试将值存储到我的数据库中。这是我用来从我的网站插入代码的代码：
<?php
$ident = $_POST['ident'];
$dato = $_POST['dato'];
$kundensnavn = $_POST['kundensnavn'];
$gsm = $_POST['gsm'];
$fodselsdato = $_POST['fodselsdato'];
$prisplan = $_POST['prisplan'];
$operator = $_POST['operator'];
$portering = $_POST['portering'];
$epost = $_POST['epost'];
// Database connection
$conn = new mysqli('localhost','my_username','my_password','id14293554_rw2');
if($conn->connect_error){
echo "$conn->connect_error";
die("Connection Failed : ". $conn->connect_error);
} else {
$stmt = $conn->prepare("INSERT INTO sales_table(ident_column, date_column, name_column, gsm_column, birthdate_column, pp_column, carrier_column, transfer_column, email_column) values(?, ?, ?, ?, ?, ?, ?, ?, ?)");
$stmt->bind_param("sssssssss", $ident, $dato, $kundensnavn, $gsm, $fodselsdato, $prisplan, $operator, $portering, $epost);
$execval = $stmt->execute();
echo $execval;
echo "Done!";
$stmt->close();
$conn->close();
}
?>
如果需要的话，这是我的index.html：
<!DOCTYPE html>
<html >
<head>
<meta charset="UTF-8">
<title>Test</title>
<link rel="stylesheet" href="./style.css">
</head>
<body>
<!-- partial:index.partial.html -->
<div class="container">
<form id="contact" action="insert.php" method="post">
<h3><center>Reg</center></h3>
<fieldset>
<input type="text" placeholder="Ident" name="ident" required autofocus>
</fieldset>
<fieldset>
<input type="text" placeholder="Dato" name="dato" required>
</fieldset>
<fieldset>
<input type="text" placeholder="Kundens navn" name="kundensnavn" required>
</fieldset>
当我按下“提交”按钮时，该脚本将运行，并给出“完成！”，因此没有错误。但是，当我检查表时，即使我输入了所有需要的信息，那里也什么也没有：
繁星淼淼1回答
-
阿晨1998
您的表包含Integer列gsm_column 但您string在此处传递值：$stmt->bind_param("sssssssss", $ident, $dato, $kundensnavn, $gsm, $fodselsdato, $prisplan, $operator, $portering, $epost); <---------尝试改成这样：$stmt->bind_param("sssisssss", $ident, $dato, $kundensnavn, $gsm, $fodselsdato, $prisplan, $operator, $portering, $epost);我也会改变最后一部分if ($stmt->execute()) { // it worked echo "Done";} else { // it didn't echo "Not inserted"; //Print error echo $stmt->error;}$stmt->close();$conn->close();
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