Tom*_*dil 5 spring spring-mvc thymeleaf spring-boot
我正在使用 spring 在 Java 中开发一个 Web 应用程序。
该应用程序包括 JavaScript 中的 Ajax 调用,该调用请求 html 代码,然后将其插入到 html 文档中。
为了将 thymeleaf 模板处理为字符串,我使用 TemplateEngine process(..) 方法。
当 thymeleaf 模板包含表单时,我遇到错误。
我的示例代码:
表单.html:
<form th:object="${customer}" xmlns:th="http://www.w3.org/1999/xhtml">
<label>Name</label>
<input type="text" th:field="*{name}" />
</form>
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AjaxController.java:
package project;
import com.fasterxml.jackson.core.JsonProcessingException;
import com.fasterxml.jackson.databind.ObjectMapper;
import org.springframework.beans.factory.annotation.Autowired;
import org.springframework.stereotype.Controller;
import org.springframework.ui.Model;
import org.springframework.web.bind.annotation.GetMapping;
import org.springframework.web.bind.annotation.ResponseBody;
import org.thymeleaf.TemplateEngine;
import org.thymeleaf.context.Context;
@Controller
public class AjaxController {
@Autowired
private TemplateEngine templateEngine;
private ObjectMapper objectMapper = new ObjectMapper();
@ResponseBody
@GetMapping(value="/form1")
public String form1() throws JsonProcessingException {
Customer customer = new Customer("Burger King");
Context templateContext = new Context();
templateContext.setVariable("customer", customer);
AjaxResponse response = new AjaxResponse();
response.html = templateEngine.process("form", templateContext);
response.additionalData = "ab123";
return objectMapper.writeValueAsString(response);
}
@GetMapping(value="/form2")
public String form2(Model model) throws JsonProcessingException {
Customer customer = new Customer("Burger King");
model.addAttribute("customer", customer);
return "form";
}
class Customer {
private String name;
public Customer(String name) {
this.name = name;
}
public String getName() {
return name;
}
public void setName(String name) {
this.name = name;
}
}
class AjaxResponse {
public String html;
public String additionalData;
}
}
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form1 是崩溃的,我试图返回 thymeleaf 模板解析的 html 代码,并在此 json 响应中包含其他数据。它在 templateEngine.process("form", templateContext); 行崩溃
将 form.html 替换为以下内容时,form1 可以工作:
客户名称是:[[${customer.name}]]
这让我得出结论,是表单标签和 th:object 导致了崩溃。
form2 按预期工作,但没有任何方法来操纵 thymeleaf 返回值。证明thymeleaf模板本身是有效的。
整个错误输出有点太大,无法粘贴到此处,但是:
org.thymeleaf.exceptions.TemplateInputException: An error happened during template parsing (template: "class path resource [templates/form.html]")
Caused by: org.thymeleaf.exceptions.TemplateProcessingException: Cannot process attribute '{th:field,data-th-field}': no associated BindStatus could be found for the intended form binding operations. This can be due to the lack of a proper management of the Spring RequestContext, which is usually done through the ThymeleafView or ThymeleafReactiveView (template: "form" - line 3, col 21)
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我的问题是:这是 spring 框架中的错误吗?或者如果不是那么我做错了什么?
更新 1:用 th:value 替换 th:field 使其工作,似乎使用 TemplateEngine .process 时表单内的 th:field 是产生错误的原因。
更新2:好的,经过大量的侦探工作,我找到了一种可以暂时使这项工作正常进行的方法。问题是 thymeleaf 需要 IThymeleafRequestContext 来处理带有表单的模板,当 TemplateEngine .process 运行时,将不会创建它。可以将其注入到您的模型中,如下所示:
@Autowired
ServletContext servletContext;
private String renderToString(HttpServletRequest request, HttpServletResponse response, String viewName, Map<String, Object> parameters) {
Context templateContext = new Context();
templateContext.setVariables(parameters);
RequestContext requestContext = new RequestContext(request, response, servletContext, parameters);
SpringWebMvcThymeleafRequestContext thymeleafRequestContext = new SpringWebMvcThymeleafRequestContext(requestContext, request);
templateContext.setVariable("thymeleafRequestContext", thymeleafRequestContext);
return templateEngine.process(viewName, templateContext);
}
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现在你可以像这样使用这个方法:
@ResponseBody
@GetMapping(value="/form1")
public String form1(HttpServletRequest request, HttpServletResponse response) throws JsonProcessingException {
Customer customer = new Customer("Burger King");
BindingAwareModelMap bindingMap = new BindingAwareModelMap();
bindingMap.addAttribute("customer", customer);
String html = renderToString(request, response, "form", bindingMap);
AjaxResponse resp = new AjaxResponse();
resp.html = html;
resp.additionalData = "ab123";
String json = objectMapper.writeValueAsString(resp);
return json;
}
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我不会将此作为答案,因为我看不出有任何理由以这种方式使用它。我正在与 Spring 人员进行沟通,以寻求真正的解决方案。
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