我將一些數據發佈到使用soome html和php的mysql數據庫中。我無法發佈收音機。我認爲我搞亂了我在php和數據庫中聲明的方式。我已將收音機定義爲tinyint。這是我的PHP單選按鈕選項不在MYSQL中發佈
$value = $_POST['name'];
$value2 = $_POST['email'];
$value3 = $_POST['address'];
$value4 = $_POST['city'];
$value5 = $_POST['state'];
$value6 = $_POST['zip'];
$value7 = $_POST['primary'];
$sql = "INSERT INTO demo (name, email, address, city, state, zip, primary) VALUES ('$value', '$value2', '$value3', '$value4', '$value5', '$value6', '$value7')";
HTML
<div id="options">
<div class="opt1"><input type="radio" name="primary" value="Color" /></div>
<div class="opt2"><input type="radio" name="primary" value="Dry" /></div>
<div class="opt3"><input type="radio" name="primary" value="Damaged" /></div>
<div class="opt4"><input type="radio" name="primary" value="Thinning" /></div>
</div>
我有其他形式的元素作爲文本字段,他們會雖然
SQL注入。 SQL注入。 <3 – 2013-04-04 01:52:25