Question

我尝试调用rest函数，但显示如下错误：
GET http://localhost:9000/api/filess/[object%20Object] 400 (Bad Request)。
可能无法获得报表数据变量的名称。我怎么得到它？

我该如何解决这个问题？

下面的组件侧：

showReport(filename: string) {
    this.reportData = new ReportData();
    this.reportData.month=this.selectedMonth;
    this.reportData.year= this.selectedYear;
    this.reportData.fileName= filename;
    var reportData: ReportData= this.reportData;
    this.defReportService.downloadFile(reportData).subscribe(
        res => {
            // var file = new Blob([res], {type: 'application/pdf'});
            var file = new Blob([res], { type: 'application/vnd.ms-excel' });
            var fileURL = URL.createObjectURL(file);
            // window.open(fileURL);
            const fname = filename + '_' + 
fileURL.toString().substr(fileURL.toString().lastIndexOf('/') + 1) + '.xls';
            saveAs(res, fname);
        },
        error => {
            console.log(error);
        }
    );
}

下面的服务端：

public downloadFile(req?: ReportData) {
    const encodedAuth = window.localStorage.getItem('encodedAuth');
    return this.http
        .get(`${this.resourceDownload}/${req}`, {
            headers: new HttpHeaders({
                Authorization: 'Basic ' + encodedAuth,
                'Content-Type': 'application/octet-stream'
            }),
            responseType: 'blob'
        })
        .pipe(
            tap(
                // Log the result or error
                data => console.log('You received data'),
                error => console.log(error)
            )
        );
}

REST下方：

@GetMapping("/filess/{reportData}")
@ResponseBody
public ResponseEntity<Resource> getFile(@PathVariable ReportData reportData) throws Exception {

    EnumList.JRList reportEnum = EnumList.JRList.valueOf(reportData.getFileName());
    File createdFile = jasperService.downloadReportFile(reportEnum, EnumList.JRExport.XLSX, reportData);
    Resource resourceFile = jasperService.loadFile(createdFile.getPath());
    String filePath = createdFile.getPath();
    //createdFile.delete();
    return ResponseEntity.ok()
        .header(HttpHeaders.CONTENT_DISPOSITION, "attachment; filename=\"" + filePath + "\"")
        .body(resourceFile);
}

Answer 1

如果您要发送对象之类的复杂数据结构，最好使用http POST方法

开机自检

const data = {...};
http.post(url , JSON.stringify(data));

如果使用GET方法，则可以使用params属性url?p=v&p2=v2

发送数据

const options= { params: new HttpParams().set('name', ...) }
http.get(url , options)

将对象映射到httpParams

  const p = { p1: "1", p2: "2", p3: "3" };
  const params = Object.entries(p).reduce(
    (data: HttpParams, par) => data.append(par[0], par[1]),
    new HttpParams()
  );
  console.log(params.toString()); // => "p1=1&p2=2&p3=3"
  http.get(url , {params})

如何在Angular中将变量名发送给rest函数？ 400（错误请求）。错误

1 个答案: