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[英]How to test Spring boot one to many JPA post request by curl o postman
[英]Testing Spring Boot Backend Post Request using Postman and Curl
我一直在尝试通过命令行使用 Postman 和 curl 来测试一些简单的 GET 和 POST 请求方法。
出于某种原因,当我尝试创建一个 json 文件并通过 Postman 发送它时,它会将所有数据保存到第一个变量中。
我不知道是怎么回事。 前端将通过 JSON 文件提供所有内容,所以如果这不起作用,那么我想在完成我的 controller 之前修复它。
这是我的药品 model:
@Entity
@Table(name = "pharmaceuticals")
public class Pharmaceutical {
@Id
@GeneratedValue(strategy = GenerationType.IDENTITY)
private long id;
@Column(name = "genericName")
private String genericName;
@Column(name = "brandNames")
private ArrayList<String> brandNames;
@Column(name = "strength" )
private String strength;
@Column(name = "quantity")
private Integer quantity;
@ManyToMany(fetch = FetchType.LAZY,
cascade = {
CascadeType.MERGE,
CascadeType.REFRESH
})
@JoinTable(name = "pharm_commonuses",
joinColumns = { @JoinColumn(name = "pharmaceutical_id") },
inverseJoinColumns = { @JoinColumn(name = "commonUse_id") })
private Set<CommonUse> commonUses = new HashSet<>();
public Pharmaceutical() {}
public Pharmaceutical(String genericName, ArrayList<String> brandNames, String strength,
Integer quantity) {
this.genericName = genericName;
this.brandNames = brandNames;
this.strength = strength;
this.quantity = quantity;
}
//getters and setters
这是我的 controller:
@CrossOrigin(origins = "http://localhost:8081")
@RestController
@RequestMapping("/api")
public class PharmaceuticalController {
@Autowired
PharmaceuticalRepository pharmRepository;
CommonUseRepository comRepository;
@GetMapping("/pharmaceuticals")
public ResponseEntity<List<Pharmaceutical>> getPharmaceuticals(@RequestParam(required = false) String title){
List<Pharmaceutical> pharms = new ArrayList<Pharmaceutical>();
pharmRepository.findAll().forEach(pharms::add);
return new ResponseEntity<>(pharms, HttpStatus.OK);
}
@PostMapping("/pharmaceuticals")
public ResponseEntity<Pharmaceutical> createPharmaceutical(@RequestBody String generic, ArrayList<String> brands, String strength, Integer quant, ArrayList<String> common){
Pharmaceutical newPharm = new Pharmaceutical(generic, brands, strength, quant);
for (String name: common) {
CommonUse com = new CommonUse(name);
comRepository.save(com);
newPharm.getCommonUses().add(com);
}
pharmRepository.save(newPharm);
return new ResponseEntity<>(newPharm, HttpStatus.CREATED);
}
}
任何帮助都会很棒!
这是你的问题:
@RequestBody String generic
你是说进来的身体应该放在这个字符串中。
您应该构建一个 object 表示您要发送的正文并将其更改为:
@RequestBody PharmaceuticalRequest generic
然后删除createPharmaceutical
function 中的所有其他输入参数。
参考: https://www.baeldung.com/spring-request-response-body#@requestbody
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