Post表单使用Fetch和检查状态OK

Question

如何获得响应代码时，张贴表单数据使用fetch香草JS，我有这段代码的帖子，但不检查响应代码与response.ok

<form method="post" id="form1" class="myForm" action="includes/vehicles/add-veh.php">
<input type="hidden" name="userid" id="userid" value="4448" />
<input type="hidden" name="vehid" id="vehid" value="84488" />
<button type="submit" name="submit" ><i class="fa-solid fa-circle-plus"></i> Save To Your List</button>
</form>

<script>
document.addEventListener('DOMContentLoaded', function() {
document.querySelector('.myForm').addEventListener('submit', function (event) {
var data = this;
fetch(data.getAttribute('action'), {
method: data.getAttribute('method'),
body: new FormData(data)
})
.then(res=>res.text())
.then(function (data) {

})
.then((res) => {
if (response.ok) { 
alert('success');
}
};

event.preventDefault();
});
});
</script>

Answer 1

这似乎可行

document.addEventListener('DOMContentLoaded', function() {
document.querySelector('.myForm').addEventListener('submit', 
function (event) {
var data = this;
fetch(data.getAttribute('action'), {
method: data.getAttribute('method'),
body: new FormData(data)
})
.then(function (response) {
if (response.ok) { 
alert('success');
} else {
alert('error');
}
})
event.preventDefault();
});
});