前端json数组对象字符串传给后端，后端如何接收保存数据库

请问这种数组对象格式的前端传到后端，后端如何接收啊。然后保存到数据库要什么格式啊。

我现在后端是这样的，那个TestQuestions字段就是来接收前端试题的。用string格式的一致报错说：Invalid character found in the request target（在请求目标中发现无效字符），请问如何处理呢。我数据库用的text格式接收的。因为题目可能比较多，我准备把所有题目一起存在一个字段内

@RestController
@RequestMapping("/TestQuestions")
public class TestQuestionsController {
    @Resource
    TestQuestionsMapper testQuestionsMapper;
    //保存词条，试卷标题，试卷内容，状态,创建人
    @CrossOrigin(origins ="*",maxAge = 3600)
    @GetMapping("/setTestQuestions")
    public void findContent(@RequestParam("entryName") String entryName, @RequestParam("paperName") String paperName, @RequestParam("paperTitle") String paperTitle, @RequestParam("state") String state,@RequestParam("TestQuestions") String TestQuestions){
        testQuestionsMapper.save(entryName,paperName,paperTitle,state,TestQuestions);
    }
}

保存的情况一般用post，接实体，你的问题都是json了也就是不固定了，那就直接一个string接吧

    @PostMapping("/setTestQuestions")
    public void findContent(@RequestBody Questions questions){
        testQuestionsMapper.save(questions);
        //testQuestionsMapper.save(questions.getEntryName(),questions.getPaperName(),questions.getPaperTitle(),questions.getStatus(),questions.getTestQuestions());
    }
public class Questions {

    private String entryName;
    private String paperName;
    private String paperTitle;
    private String status;
    private String testQuestions;

    public String getEntryName() {
        return entryName;
    }

    public void setEntryName(String entryName) {
        this.entryName = entryName;
    }

    public String getPaperName() {
        return paperName;
    }

    public void setPaperName(String paperName) {
        this.paperName = paperName;
    }

    public String getPaperTitle() {
        return paperTitle;
    }

    public void setPaperTitle(String paperTitle) {
        this.paperTitle = paperTitle;
    }

    public String getStatus() {
        return status;
    }

    public void setStatus(String status) {
        this.status = status;
    }

    public String getTestQuestions() {
        return testQuestions;
    }

    public void setTestQuestions(String testQuestions) {
        this.testQuestions = testQuestions;
    }
}

搞个VO啊，List

你把你要接收的参数整合到一个类当中，比如A，然后用List<A>接收