我是非常新的php mysql，我正在使用下面的php从输入数据.我的var_dump显示所有的值，我已经正确地输入在表单中.数据库连接也是成功的.但是数据没有插入到MariaDB表.我已经交叉检查了表结构，值和数据类型是匹配的.一切看起来都是好的，但是数据库表从来没有用数据更新过.任何帮助都将不胜感激！提前感谢.
dropdown-form.php
=================
<?php
include('database.php');
?>
<!DOCTYPE html>
<html>
<head>
<meta charset="utf-8">
<meta name="viewport" content="width=device-width, initial-scale=1">
<title>Chiefdom Zambia</title>
<link rel="stylesheet" href="https://cdn.jsdelivr.net/npm/water.css@2/out/water.min.css">
<style>
.container
{
border-radius: 5px;
background-color: #f2f2f2;
padding: 20px;
}
.col-25
{
float: left;
width: 25%;
margin-top: 6px;
}
.col-75
{
float: left;
width: 75%;
margin-top: 6px;
}
/* Clear floats after the columns */
.row:after {
content: "";
display: table;
clear: both;
}
/* Responsive layout - when the screen is less than 600px wide, make the two columns stack on top of each other instead of next to each other */
@media screen and (max-width: 600px) {
.col-25,
.col-75,
input[type=submit] {
width: 100%;
margin-top: 0;
}
}酋长领地授权数据收集
<div class="container">
<div class="dependent-dropdown">
<form method="post" action="dropdown-form.php">
<legend style="text-align: center"> Land Info </legend>
<legend style="text-align: center"> Country : Zambia </legend>
<div class="input-field">
<select name="pro_id" id="country">
<option value="pro_id">Select your Province here</option>
<?php
$countryData = "SELECT id, name from countries";
$result = mysqli_query($conn, $countryData);
if (mysqli_num_rows($result) > 0) {
while ($arr = mysqli_fetch_assoc($result)) {
?>
<option value="<?php echo $arr['id']; ?>">
<?php echo $arr['name']; ?>
</option>
<?php
}
}
?>
</select>
</div>
<div class="input-field">
<select name="dist_id" id="state">
<option value="dist_id">Select your District here</option>
</select>
</div>
<div class="input-field">
<select name="chief_id" id="city">
<option value="chief_id">Select your Chiefdom here </option>
</select>
</div>
<input type="text" placeholder="Enter Zone Number " name="zone_no">
<input type="text" placeholder="Enter Zone Name " name="zone_name">
<input type="text" placeholder="Enter Locality" name="locality">
<legend style="text-align: center"> Farmer Info </legend>
<input type="text" placeholder="Enter Name of the Farmer" name="farmer_name"></input>
<input type="text" placeholder="Enter NRC Number of the Farmer " name="nrc_no">
<input type="text" placeholder="Enter Contact Number of the Farmer " name="contact_no">
<input type="text" placeholder="Land Holding in Hectare (Ha)" name="land_holding">
<legend style="text-align: center"> Crop Info </legend>
<select name="crops_no" id="form-selector">
<option value="0">Select Total Number of Crops </option>
<option value="1">1</option>
<option value="2">2</option>
<option value="3">3</option>
<option value="4">4</option>
<option value="5">5</option>
</select>
<br>
<button>submit</button>
</form>
</div>
</div>
</body>
</html>
<?php
$pro_id = filter_input(INPUT_POST, "pro_id", FILTER_VALIDATE_INT);
$dist_id = filter_input(INPUT_POST, "dist_id", FILTER_VALIDATE_INT);
$chief_id = filter_input(INPUT_POST, "chief_id", FILTER_VALIDATE_INT);
$zone_no = $_POST["zone_no"];
$zone_name = $_POST["zone_name"];
$locality = $_POST["locality"];
$farmer_name = $_POST["farmer_name"];
$nrc_no = $_POST["nrc_no"];
$contact_no = $_POST["contact_no"];
$land_holding = $_POST["land_holding"];
$crops_no = filter_input(INPUT_POST, "crops_no", FILTER_VALIDATE_INT);
ini_set('display_errors', 1); ini_set('log_errors',1); error_reporting(E_ALL); mysqli_report(MYSQLI_REPORT_ERROR | MYSQLI_REPORT_STRICT);
$sql = "INSERT INTO farmer (' province`, `district`, `chiefdom`, `zone no`, `zone name`,`local`, `farmer name`, `nrc no`, `mobile no`, `land holding`, `no of crops`, ) VALUES ( ?,?,?,?,?,?,?,?,?,?,?)";
$stmt = mysqli_stmt_init($conn);
if (!mysqli_stmt_prepare($stmt, $sql)) {
die(mysqli_error($conn));
}
mysqli_stmt_bind_param(
$stmt,
"iiisssssssi",
$pro_id,
$dist_id,
$chief_id,
$zone_no,
$zone_name,
$locality,
$farmer_name,
$nrc_no,
$contact_no,
$land_holding,
$crops_no
);
if (mysqli_stmt_execute($stmt)) {
echo "Record added successfully";
} else {
echo "Error: " . mysqli_error($conn);
}
mysqli_stmt_close($stmt);
database.php
============
$dbHost = "localhost";
$dbUsername = "root";
$dbPassword = "";
$dbName = "ddl";
$conn = new mysqli($dbHost, $dbUsername, $dbPassword, $dbName);
if ($conn->connect_error)
{
die("Connection failed: " . $conn->connect_error);
}
if (mysqli_connect_errno()) {
die("Connection Error: " . mysqli_connect_error());
}'情况如下：
我已经检查了数据库连接的错误。数据类型和值是完美的。表列和sql查询是匹配的。sql查询的手动输入接受数据并在表中更新。
我得到了以下错误：致命错误：未捕获的mysqli_sql_异常：SQL语法中有错误;查看与您的MariaDB服务器版本对应的手册，在下拉表单中的第1行的'' province , district , chiefdom , zone no , zone name , local , farmor...'附近使用正确的语法。php：255堆栈跟踪：第0个下拉列表.php（255）：mysqli_stmt_prepare（对象（mysqli_stmt），'插入到远端...'）#1 {main}在dropdown-form.php的第255行抛出
2条答案
按热度按时间xqk2d5yq1#
看起来你在字段名周围使用了一个反勾号而不是一个直引号。我建议使用一个可以突出显示语法的编辑器（有很多免费的），这样这些问题就会跳到你的眼睛里。这会把你的注意力集中到更重要的部分。
8yoxcaq72#
这里的MySQL语法是不正确的，这正是错误发现问题的地方：
一定要 * 极其 * 密切注意MySQL在错误信息中说的话。它通常会精确地指出问题所在，就像这里所做的那样。