如果HTTP Status Code出现问题，我正在尝试更改PHP中的SQL Query，并HTTP Status Code通过AJAX来接收此jQuery registro.php }}
为什么不工作？这是<?php
require_once('conexao.php'); // the connection to db
$telefone = $_POST['telefone']; // get the tel number
$celular = $_POST['celular']; // get the cel number
$celular = str_replace(array("(",")","-"," "), "", $celular);
$dddCelular = substr($celular, 0, 2);
$celular = substr($celular, 2, 9);
$telefone = str_replace(array("(",")","-"," "), "", $telefone);
$dddTelefone = substr($telefone, 0, 2);
$telefone = substr($telefone, 2, 8);
// after split the region code I check if is equal to a specify region, in case 11
if($dddCelular == "11" || $dddTelefone == "11") {
$sql = "INSERT INTO `bigdata` (`id`, `origem`, `nome`, `dddtelefone`, `telefone`, `dddcelular`, `celular`, `email`, `obsadm`, `observacoes`, `timestampcadastro`)
VALUES (NULL, 'renovacnh', '".utf8_decode($nome)."', '".$dddTelefone."', '".$telefone."', '".$dddCelular."', '".$celular."', '".$email."', '".$horario."', '".$observacoes."', CURRENT_TIMESTAMP);";
if(mysqli_query($conn, $sql)) { mysqli_close($conn);
echo "OK";
http_response_code(200); // the query is ok! so, change to 200
};
} else {
// var_dump(http_response_code()); // int(200) - default
http_response_code(503); // set a new code
// var_dump(http_response_code()); // int(503) - new code
echo "ERROR";
}
文件内容：
AJAX这是我的<script>
function pixel(data) {
console.log(data);
if(data == "OK") {
alert("It works!");
} else {
alert("Fail");
}
}
function envia() {
$.ajax({
method: "POST",
url : "php/registro.php",
data: {
"celular": "(21) 95494-3943", // the phone is wrong region code, not 11
"telefone": "(21) 4494-3943", // the cel is wrong region code, not 11
},
success: function (data,xhr) {
pixel(data);
console.log(xhr); // trow success
},
error: function (error,xhr) {
console.log(error);
console.log(xhr);
}
});
}
envia();
</script>
脚本：
AJAX所以最后，我试图通过if()检查响应是否等于“OK”，并且HTTP Status Code以某种方式失败。并且根本没有验证为OK响应。所以我尝试使用AJAX以某种方式检查@Override
public void deleteEntry(long id) {
String deleteQuery = "DELETE FROM company WHERE id = ?";
jdbcTemplate.update(deleteQuery, id);
}
响应。
有什么想法吗？谢谢！