我已经尝试了许多不同的方法将数据插入到我的数据库中。我更进一步了，它过去只是说错误，但现在当你提交表单时，它会加载一个空白页，但表单中的数据不会添加到表中；/
<form name="datainsert" method="post" action="dataInsert.php">
<label>Server Name: </label>
<input type="text" name="name" placeholder="Enter Server Name" style="margin-left:90px; width:160px; padding:5px; margin-top:10px;"><br />
<label>Server Location:</label>
<input type="text" name="location" placeholder="Enter Server Location" style="margin-left:71px; width:160px; padding:5px; margin-top:10px;"><br />
<label>Server Operating System:</label>
<input type="text" name="os" placeholder="Enter Server OS" style="margin-left:16px; width:160px; padding:5px; margin-top:10px;"><br/>
<input style="margin-top:10px;" name="submit" value="submit" type="submit">
</form>
<?php
include 'dbconnect.php';
$name = $_POST['name'];
$location = $_POST['location'];
$os = $_POST['os'];
)
mysql_query("INSERT INTO fostvm (name, location, os) VALUES ('$name', '$location', '$os')");
$result=mysql_query($sql);
if($result){
echo "Data Added Successfully";
} else {
echo "Error";
}
?>
有人能看到语法错误吗？或者我可能在哪里出错
谢谢！
试试这个。您的代码是Ok的，只需注释这一行（$result=mysql_query（$sql）；）。使用此代码。为什么在代码中尝试了两次mysql_qury（）。
<?php
include 'dbconnect.php';
$name = $_POST['name'];
$location = $_POST['location'];
$os = $_POST['os'];
)
$result = mysql_query("INSERT INTO fostvm (name, location, os) VALUES ('$name', '$location', '$os')");
//$result=mysql_query($sql);
if($result){
echo "Data Added Successfully";
} else {
echo "Error";
}
?>
你没有$sql variable，你正在使用它mysql_query试试这个，
<?php
include 'dbconnect.php';
if(isset($_POST['submit'])){ // check for form submit
$name = $_POST['name'];
$location = $_POST['location'];
$os = $_POST['os'];
$result=mysql_query("INSERT INTO fostvm (name, location, os) VALUES ('$name', '$location', '$os')");
if($result){
echo "Data Added Successfully";
} else {
echo "Error";
}
}
?>
您还应该使用mysqli，因为mysql是deprecated