我正在尝试使用php和html表单插入mysql数据库。当我按submit时,它表示已成功插入项目,但当我登录到phpmyadmin时,表为空?
我的html表单代码是 creon.html
.
<form name="bmr_calc" action="https://cs1.ucc.ie/~lmm12/Project/creon.php" method="POST" id="BMRform">
<h1 id="info">Creon Calculator</h1>
<table>
<tr>
<td colspan="2" align="center">I take one: <br>
<input type="radio" name="creon1" value="10" required> Creon 10,000
<input type="radio" name="creon1" value="25" required> Creon 25,000
<br>per <input type="text" name="creon3" required>g of fat
</td>
</tr>
<br>
<tr>
<td>There are:</td>
<td><input type="text" name="creon4" required>g of fat in the item I'm about to eat</td>
</tr>
</table>
<td><input type="submit" value="Submit" style="float:center">
<br>
<img src="img/regpic.jpg" alt="reg" id="reg">
<br>
</td>
</form>
我的php代码是 creon.php
它保存在我的大学服务器上;
<?php
$creon1 = $_POST['creon1'];
$creon3 = $_POST['creon3'];
$creon4 = $_POST['creon4'];
if (!empty($creon1) || !empty($creon3) || !empty($creon4)) {
$host = "cs1.ucc.ie";
$dbUsername = "lmm12";
$dbPassword = "----";
$dbname = "mscim2018_lmm12";
//create connection
$conn = new mysqli($host, $dbUsername, $dbPassword, $dbname);
if (mysqli_connect_error()) {
die('Connect Error('. mysqli_connect_errno().')'. mysqli_connect_error());
} else {
$INSERT = "INSERT Into creon (creon1, creon3, creon4) values(?, ?, ?)";
//Prepare statement
$stmt = $conn->prepare($INSERT);
$stmt->bind_param("sss", $creon1, $creon3, $creon4);
$stmt->execute();
echo "New record inserted sucessfully";
$stmt->close();
$conn->close();
}
} else {
echo "All field are required";
die();
}
1条答案
按热度按时间f3temu5u1#
这似乎要么是mysql中的类型问题,要么是php代码中的类型问题
当然,从creon表上传数据类型是varchars、text等字段。
请注意bind_param周围的if/else语句,您也需要这样做,以防出现不太正确的情况,请在语句中大写into。
我运行了以下命令:
它在提交表格后给了我这个结果:
在我的数据库中,它插入了以下行: