我的UPDATE查询失败了，尽管语法对我来说很好（我有另一个更新查询在同一页面上运行良好）。
mysql_connect("localhost", "root", "") or die(mysql_error());
mysql_select_db("sitename") or die(mysql_error());
$id = $_GET['id'];
if (isset($_POST['submit'])){
$b = mysql_real_escape_string(trim($_POST['body']));
//**You have an error in your SQL syntax;** --> ?
mysql_query ("UPDATE body SET body= $b WHERE id = $id") or die (mysql_error() );
// $b is fine
echo "$b";
}
HTML审阅表单的呈现方式。。
// Puts SQL Data into an array
$q = mysql_query("SELECT * FROM vote") or die (mysql_error());
// Now we loop through the database
echo "<br />";
while ($ratings = mysql_fetch_array($q))
{
//This outputs the doctors's name
echo "Doctor's name:" . $ratings['doctor_name'] ."<br />";
//This outputs a textarea for the user to submit comments
echo "<b>Your Experience: </b>";
echo "<form method='post' action='review_doctors.php'>
<textarea name='body'></textarea>
<input type='submit' name='submit' value='Send' id='submit'/>
</form>
";
echo "<br />";
echo "<p> </p>";
}
为什么每当提交评论时都会出现SQL语法错误？
因此，您正在从$_GET数组中设置$id，这可能不会在通过post提交表单时设置。
您正在运行的更新查询在POST检查中（检查是否设置了$_POST['submit']）。
您可能想要在post正文中发送$id的值，并从post数组中提取它。
我将其修复为：
// If submitted
if (isset($_POST['id'])){
//Capture what was typed in textarea
$b = mysql_real_escape_string(trim($_POST['body']));
$id = $_POST['id'];
mysql_query ("UPDATE vote SET body = '$b' WHERE id = $id") or die (mysql_error() );
// $b and $id are still fine
echo "$b";
echo "$id";
}
还修复了隐藏的输入值：
while ($ratings = mysql_fetch_array($q))
{
//This outputs the doctors's name
echo "Doctor's name:" . $ratings['doctor_name'] ."<br />";
$id = $_POST['id'];
//This outputs a textarea for the user to submit comments
echo "<b>Your Experience: </b>";
echo "<form method='post' action='review_doctors.php'>
<textarea name='body'></textarea>
<input type='submit' name='submit' value='Send'/>
<input type='hidden' name='id' value='$ratings[id]' />
</form>
";
echo "<br />";