我有一个
HTML格式的表单,但它的设置并不像我想的那样正常.我将发布代码形式的HTML(
PHP)文件和
PHP将其发送到数据库.
<form action="upload.PHP" method="POST"> <!-- Name input--> <div class="form-group"> <label class="control-label" for="name">Name</label> <div class=""> <input id="name" name="name" placeholder="First and Last Name" class="form-control input-md" type="text" required> </div> </div> <div class="form-group"> <label class=" control-label" for="supportingDoc">Upload Supporting Documentation</label> <div class=""> <input id="supportingDoc" name="supportingDoc" class="input-file" type="file" style="margin-top: .5em; margin-left: 4em;"> </div> </div> <hr> <!-- Submit --> <div class="form-group"> <label class="control-label" for="submit"></label> <div class=""> <button value="Submit" type="submit" id="submit" name="submit" class="btn btn-danger" style="border-radius: 25px;">Submit</button> </div> </div> </form>
<?PHP $servername = "localhost"; $username = "xxx"; $password = "xxx"; $dbname = "xxx"; // Create connection $con = MysqLi_connect("localhost","xxx","xxx"); // Check connection if (MysqLi_connect_errno()){ echo "Failed to connect to MysqL: " . MysqLi_connect_error(); } if (isset($_REQUEST['name'])){ // set variables $name = MysqL_real_escape_string($_POST['name']); $supportingDoc = MysqL_real_escape_string($_POST['supportingDoc']); $sql = "INSERT INTO `tablew` (name,supportingDoc) VALUES ('$name','$supportingDoc')"; $result = MysqLi_query($con,$sql); if ($con->query($sql) === TRUE) { echo "New record created successfully"; printf("New record created successfully"); } else { echo "Error: " . $sql . "<br>" . $con->error; printf("Error: " . $sql . "<br>" . $con->error); } $con->close(); } ?>
我尝试了各种各样的变化,在PHPmyadmin中没有显示任何内容.我甚至从我之前创建的网站复制了它仍然没有工作大声笑.
我看到有登录信息的变量并且仍然把它放在MysqLi中,但是我现在已经在这一件事情大约8个小时了,并且没有果汁,所以希望有人看到我搞砸了.
在此先感谢大家的帮助.
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更新:
我做了上面提到的所有更改,现在得到这个:
警告:MysqLi_connect():( HY000 / 1049):未知数据库
我可以在PHPmyadmin和Sequel Pro中看到数据库.我还确保设置密码并登录“root”.我的代码如下登录:
$con = MysqLi_connect("localhost","root","epboarding"); // Check connection if (MysqLi_connect_errno()) { echo "Failed to connect to MysqL: " . MysqLi_connect_error(); }
这是我的帖子:
if (isset($_REQUEST['submit'])){ // set variables $name = MysqLi_real_escape_string($con,$_POST['name']); $sql = "INSERT INTO `wos` (name) VALUES ('$name')"; $result = MysqLi_query($con,$sql);