如何处理POST请求中的变量?假设我有这样的表,我想更新特定id所在的投票变量。
所以对于代码,我有这个
$vote = $_POST["votes"];
$sentid = $_POST["sentid"];
以及类似的东西
UPDATE `my_exampleo202s`.`President Candidates` SET votes = votes + 1 WHERE `President Candidates`.`id` = sentid
在邮件请求中,我给sentid发了一个号码。sentid=3
但这不起作用。我不能给出任何错误或任何东西,因为我不是从浏览器中查看的。
你知道我应该怎么做吗?
(如果需要的话,这是我现在的代码)
<?php
$vote = $_POST["votes"];
$sentid = $_POST["sentid"];
$conn = new mysqli("localhost","exampleo202s","","my_exampleo202s");
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
$sql = "UPDATE `my_exampleo202s`.`President Candidates` SET votes = votes + 1 WHERE `President Candidates`.`id` = sentid";
if ($conn->query($sql) === TRUE) {
echo "<br>Record updated successfully <br>";
} else {
echo "<br>Error updating record: <br>" . $conn->error;
}
$conn->close();
?>
更新
<?php
$vote = $_POST["votes"];
$sentid = $_POST["sentid"];
$conn = new mysqli("localhost","jusavoting10rxx9s3","","my_jusavoting10rxx9s3");
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
$sql = "UPDATE `my_jusavoting10rxx9s3`.`President Candidates` SET votes = votes + 1 WHERE `President Candidates`.`id` = $sentid";
if ($conn->query($sql) === TRUE) {
echo "<br>Record updated successfully <br>";
} else {
echo "<br>Error updating record: <br>" . $conn->error;
}
$conn->close();
?>
如果$_POST值不起作用,请尝试回显:
<?php
$vote = $_POST["votes"];
$sentid = $_POST["sentid"];
$conn = new mysqli("localhost","exampleo202s","","my_exampleo202s");
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
$sql = "UPDATE `my_exampleo202s`.`President Candidates` SET votes = votes + 1 WHERE `President Candidates`.`id` = $sentid";
if ($conn->query($sql) === TRUE) {
echo "<br>Record updated successfully <br>";
} else {
echo "<br>Error updating record: <br>" . $conn->error;
}
$conn->close();
?>