所以我试图根据用户输入调用一个值。我已经为表单编写了代码,但我需要根据用户在表单上键入的内容从服务器调用值
这是我迄今为止写的:
@$conn = new mysqli($host, $user, $pass, $db_name);
$resault = $conn->query("SELECT * FROM field_data_field_paint_efficeincy_");
$rows=$resault->fetch_all(MYSQLI_ASSOC);
echo '<pre>',print_r($rows),'</pre>';foreach($rows as $row){
echo $row['field_paint_efficeincy__value'], '<br>';
}
?>
<form action="" method="post">
Variable: <input type="number" name="P" value="0" /> <br/>
类似这样的东西:
<?php
if(isset($_GET["P"])) {
$conn = new mysqli($host, $user, $pass, $db_name);
$resault = $conn->query("SELECT * FROM field_data_field_paint_efficeincy_ WHERE field_paint_efficeincy__value = '" . $_GET["P"] . "'");
//Assuming field_paint_efficeincy__value is the column that should match the input value
if ($resault->num_rows > 0) {
while($row = $resault->fetch_assoc()) {
echo $row['field_paint_efficeincy__value'] . "<br />";
}
}
$conn->close();
}
?>
<form action="<?php echo $_SERVER['PHP_SELF']?>" method="get">
Variable: <input type="number" name="P" value="0" /> <br/>
<input type="submit" value="Submit" name="submitButton" />
</form>
您是否试图在查询中使用表单提交作为动态变量?如果是这样,您需要获取已发布的值,并将其分配为要在查询中使用的变量。请记住对您提交的帖子进行消毒和保护。。这不仅仅是一个例子。
<?php
$p = $_POST['P'];
$resault = $conn->query("SELECT * FROM field_data_field_paint_efficeincy_ where p='$p'");
?>
假设p是数据库列的名称。