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我无法让此表单成功将数据提交到数据库。我已经验证了数据库/表存在（如下图所示），并且由于某种原因它没有将其插入数据库。当我尝试提交示例数据时，它不会更改页面而只是坐在那里。到底是怎么回事？ 

<body>
<?php if($_SERVER["REQUEST_METHOD"] != "POST"){ ?>
<h1>My Favorite Foods</h1>

<form action="index.php" method="post" id="foodForm">
Name: <input type="text" name="foodname" id="nameField"></input><br />
Type: <select name="foodtype" id="typeField">
<option value="fruit">Fruit</option>
<option value="vegetable">Vegetable</option>
<option value="dairy">Dairy</option>
<option value="meat">Meat</option>
<option value="grain">Grain</option>
<option value="other">Other</option>
</select><br />
Number of Calories: <input type="text" name="foodcals" id="calsField"></input><br />
Healthy? <input type="checkbox" name="foodhealth" value="healthy" id="healthyField"></input><br />
Additional Notes:<br />
<textarea name="foodnotes" id="notesField"></textarea><br />
<input type="submit" value="Add" onclick="validateForm();return false;"></input>
</form>

<?php }else{ ?>
<!-- form handling and output printing stuff goes here -->
<?php $insert = "INSERT INTO Foods(Name, Type, NumCals, Healthy, Notes) VALUES ('$_POST[foodname]', '$_POST[foodtype]', '$_POST[foodcals]', '$_POST[foodhealth]', '$_POST[foodnotes]'"; 
$con = mysqli_connect("localhost","root");
if (mysqli_connect_errno())
{
    echo "Failed to connect to MySQL: " . mysqli_connect_error();
}
mysqli_select_db("mydb"); 
$result = mysqli_query($con, $insert); 
if ($result) { 
    echo "Food added successfully."; 
    /*while ($row = mysqli_fetch_array($result)) { 
        echo $row['Name'] . ", " . $row['Type'] . ", " . $row['NumCals'] . ", " . $row['Healthy'] . ", " . $row['Notes']; 
        echo "<br>"; 
     } */
} else { 
     echo "Error adding person";  
     mysqli_error($con); 
} 
} ?>
</body>
</html>

架构：

食品（PID INT NOT NULL AUTO_INCREMENT,PRIMARY KEY(PID),Name VARCHAR(20),Type VARCHAR(9),NumCals INT,Healthy BOOL,Notes TEXT）

4

1 回答 1

3

提交按钮中的这个属性：

onclick="validateForm();return false;"

阻止表单提交。return false意味着浏览器不应该通过单击按钮执行默认操作。

假设validateForm()函数返回一个指示验证是否成功的布尔值，请将其更改为：

onclick="return validateForm();"

改变：

} else { 
     echo "Error adding person";  
     mysqli_error($con);

至：

} else { 
     echo "Error adding person: " . mysqli_error($con);

这样错误信息就会显示出来。

而你在你的陈述)结束时错过了。INSERT

于 2013-10-06T01:19:48.367 回答