java.util.MissingFormatArgumentException: 格式说明符“%s”
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java.util.MissingFormatArgumentException: Format specifier '%s'
提问by Leon
I'm obviously missing something but I don't know what...
It's stupid how much little things make you go crazy more than complicate ones...
This is the controller's code:
我显然遗漏了一些东西，但我不知道是什么......
愚蠢的小事情比复杂的事情更让你发疯......
这是控制器的代码：
@RequestMapping(value = "/getClienteNomeCognome", method = RequestMethod.GET)
public ResponseEntity<List<Object>> getClienteNomeCognome(@RequestParam("nomeCliente") String nomeCliente,
@RequestParam("cognomeCliente") String cognomeCliente) {
List<Object> listaRisultati = new ArrayList<Object>();
try {
listaRisultati = serviziDocumentaleService.getClienteNomeCognome(nomeCliente, cognomeCliente);
} catch (Exception e) {
LOGGER.warn(String.format("Errore inatteso sulla chiamata del servizio: [%s]", e.toString()));
}
LOGGER.info(String.format("Avvio ricerca cliente con nome: %s, cognome: %s)", nomeCliente, cognomeCliente));
return new ResponseEntity<List<Object>>(listaRisultati, HttpStatus.OK);
}
And this is getClienteNomeCognome:
这是 getClienteNomeCognome：
public List<Object> getClienteNomeCognome(String nome, String cognome) throws Exception {
try {
final RestTemplate restTemplate = new RestTemplate();
final String url = "somelink?cognome=%25"+cognome+"%25&nome=%25"+nome+"%25";
final ResponseEntity<List> response = (ResponseEntity<List>) restTemplate.getForObject(url, List.class);
if (response.getBody() != null && response.getBody().toString().contains("<error>")) {
throw new Exception(String.format(
"La risposta del servizio contiene degli errori: %s",
response.getBody()));
} else {
LOGGER.debug("Fine chiamata al servizio di ricerca cliente");
return response.getBody();
}
} catch (HttpClientErrorException hcee) {
throw new Exception(String.format(
"Errore durante la chiamata. Error: %s",
hcee.getMessage()));
} catch (Exception e) {
throw new Exception(String.format(
"Errore generico durante la chiamata al servizio. Error: %s"
+ e.getMessage()));
}
}
回答by McDowell
throw new Exception(String.format(
"Errore generico durante la chiamata al servizio. Error: %s"
+ e.getMessage()));
should be
应该
throw new Exception(String.format(
"Errore generico durante la chiamata al servizio. Error: %s",
e.getMessage()));
回答by DGonz
I can't be 100% sure of the issue, since its missing code so I can emulate it from here. But it looks like:
我不能 100% 确定这个问题，因为它缺少代码，所以我可以从这里模拟它。但它看起来像：
LOGGER.info(String.format("Avvio ricerca cliente con nome: %s, cognome: %s)", nomeCliente, cognomeCliente));
has an extra )after the last %s, so maybe its just not reading it correctly? Unless that's just a mistake when copy-pasting the code here.
)在最后一个之后有一个额外的%s，所以也许它只是没有正确阅读它？除非在此处复制粘贴代码时这只是一个错误。
Let us know if that works.
让我们知道这是否有效。
