xmlhttp 傳遞變量消失_如何使用JavaScript XMLHttpRequest将多個參數傳遞給@RequestBody

function saveSchemaInDatabase(schemaName) {

var xhttp = new XMLHttpRequest();

xhttp.onreadystatechange = function() {

};

xhttp.open("POST", "/user/saveSchemaInDatabase", true);

xhttp.send(schemaName);

}

我用這種方式在我的控制器中捕捉到那一個鏡頭:

@PostMapping(path = { "/user/saveSchemaInDatabase" })

public String saveSchemaInDatabase(@RequestBody String schemaName) {

return "redirect:/user";

}

//shoot

function saveSchemaInDatabase(schemaName, diagramJson) {

var xhttp = new XMLHttpRequest();

xhttp.onreadystatechange = function() {

};

xhttp.open("POST", "/user/saveSchemaInDatabase", true);

xhttp.send(schemaName, diagramJson);

}

//catch

@PostMapping(path = { "/user/saveSchemaInDatabase" })

public String saveSchemaInDatabase(@RequestBody String schemaName, @RequestBody String diagramJson) {

return "redirect:/user";

}

我希望你知道我的意思。當然我的方法行不通。出現錯誤400。

有人能幫我嗎?即時消息完成:(