当我尝试将数据插入我的数据库时,它无法正常工作,当我尝试发送数据时也没有显示任何错误。 这是HTML表单
<form method="post" action="check_inschrijven.php" id="inschrijf_form">
<label>Naam</label><br>
<input type="text" name="naam" class="inputs_text"><br>
<label>Cursus</label><br>
<input type="text" name="cursus" class="inputs_text"><br>
<label>E-Mail</label><br>
<input type="text" name="email" class="inputs_text"><br><br>
<input type="submit" value="Registreren" name="reg_button" class="reg_button">
</form>
至于PHP代码,它就在这里:
<?php
$link = mysql_connect("localhost", "uname", "password", "mkb");
if(!$link)
{
die('Kon niet verbinden met de database: ' . mysql_error());
}
$naam = $_POST['naam'];
$cursus = $_POST['cursus'];
$email = $_POST['email'];
$query = "INSERT INTO cursist (naam, cursus, email) VALUES ('$naam', '$cursus', '$email')";
if(!mysql_query($query))
{
die("Gefaald om gegevens op te slaan");
header("location: inschrijven.php");
} else {
echo "Gegevens succesvol toegevoegd aan de database";
header("location: index.php");
}
mysql_close();
?>
答案 0 :(得分:0)
您可以修改mysql_query。您可以提供Db参考并尝试它。尝试:
if (!mysql_query($query, $link)) {
// logic
}
希望这有帮助。
答案 1 :(得分:0)
试试这段代码
$query = mysql_query("INSERT INTO cursist (naam, cursus, email) VALUES ('$naam', '$cursus', '$email')");
if(!($query))
{
die("Gefaald om gegevens op te slaan");
header("location: inschrijven.php");
} else {
echo "Gegevens succesvol toegevoegd aan de database";
header("location: index.php");
}
答案 2 :(得分:0)
找到解决方案,这有效:
<?php
$link = mysqli_connect("localhost", "root", "usbw", "mkb");
if($link === false){
die("ERROR: Could not connect. " . mysqli_connect_error());
}
$naam = mysqli_real_escape_string($link, $_REQUEST['naam']);
$cursus = mysqli_real_escape_string($link, $_REQUEST['cursus']);
$email = mysqli_real_escape_string($link, $_REQUEST['email']);
$sql = "INSERT INTO cursist (naam, cursus, email) VALUES ('$naam', '$cursus', '$email')";
if(mysqli_query($link, $sql)){
echo "Succesvol ingeschreven";
header("location: index.php");
} else{
echo "ERROR: Could not able to execute $sql. " . mysqli_error($link);
}
mysqli_close($link);
?>