Question

我有2个模型类。律师和任命。

律师模型

int id;
String firstName;
String gender;
String school;

Appontment Model

int id;
String appDate;
String appTime;
Lawyer lawyer;

首先，我要展示律师名单（这是视图）。律师名单应显示律师的详细信息（姓名，年龄，学校和ALSO 预约时间和日期）。

如何加入这两个模型并在一个视图中显示两个记录

查看

我认为View中存在问题。最初我使用Scafolding生成视图。这会导致问题吗？

<%@ page import="Pro.Appointment"%>

<%@ page import="Pro.Lawyer"%>
<!DOCTYPE html>
<html>
<head>
<meta name="layout" content="main">
<g:set var="entityName"
    value="${message(code: 'appointment.label', default: 'Appointment')}" />

<g:set var="entityName"
    value="${message(code: ' lawyer.label', default: ' Lawyer ')}" />
<title><g:message code="default.show.label" args="[entityName]" /></title>
</head>
<body>
    <a href="#show-lawyerr" class="skip" tabindex="-1"><g:message
            code="default.link.skip.label" default="Skip to content&hellip;" /></a>
    <div class="nav" role="navigation">
        <ul>
            <li><a class="home" href="${createLink(uri: '/')}"><g:message
                        code="default.home.label" /></a></li>
            <li><g:link class="list" action="list">
                    <g:message code="default.list.label" args="[entityName]" />
                </g:link></li>
            <li><g:link class="create" action="create">
                    <g:message code="default.new.label" args="[entityName]" />
                </g:link></li>
        </ul>
    </div>
    <div id="show-lawyer" class="content scaffold-show" role="main">
        <h1>
            <g:message code="default.show.label" args="[entityName]" />
        </h1>
        <g:if test="${flash.message}">
            <div class="message" role="status">
                ${flash.message}
            </div>
        </g:if>
        <ol class="property-list lawyer">

            <g:if test="${lawyerInstance?.firstName}">
                <li class="fieldcontain"><span id="firstName-label"
                    class="property-label"><g:message
                            code="lawyer.firstName.label" default="First Name" /></span> <span
                    class="property-value" aria-labelledby="firstName-label"><g:fieldValue
                            bean="${lawyerInstance}" field="firstName" /></span></li>
            </g:if>

            <g:if test="${lawyerInstance?.lastName}">
                <li class="fieldcontain"><span id="lastName-label"
                    class="property-label"><g:message
                            code="lawyer.lastName.label" default="Last Name" /></span> <span
                    class="property-value" aria-labelledby="lastName-label"><g:fieldValue
                            bean="${lawyerInstance}" field="lastName" /></span></li>
            </g:if>

            <g:if test="${lawyerInstance?.gender}">
                <li class="fieldcontain"><span id="gender-label"
                    class="property-label"><g:message code="lawyer.gender.label"
                            default="Gender" /></span> <span class="property-value"
                    aria-labelledby="gender-label
<span
                    class="property-value" aria-labelledby="speciality-label"><g:fieldValue
                            bean="${lawyerInstance}" field="speciality" /></span></li>
            </g:if>



ddfdfdfd
        <g:each var="appointments" in="${lawyer}">
        sdsd
            <p>
                firstdsdsdsName:
                ${lawyer.firstName}
            </p>
            <p>
                gender:
                ${lawyer.gender}
            </p>
            <p>appointments:</p>
            <ul>
                <g:each var="app" in="${lawyer.appointments}">
                    <li>
                        ${app.appointmentTime}
                    </li>
                </g:each>
            </ul>
        </g:each>












</body>
</html>

Answer 1

首先，我建议您更进一步，以这种方式更新律师模型：

String firstName
String gender
String school

static hasMany = [appointments: Appointment]

约会模式：

Date appDate
Date appTime

static belongsTo = [lawyer: Lawyer]

使用hasMany和belongsTo属性更改默认的Grails级联行为，保存和更新，保存，更新和删除（因此，如果删除律师，每个律师的约会也会被删除）。

要显示每位律师和每位律师的预约，您可以这样做：首先在您的控制器方法中返回[lawyers: Lawyers.list()]之类的内容，然后在与此方法相关的视图中添加此代码：

<g:each var="lawyer" in="${lawyers}">
    <p>firstName: ${lawyer.firstName}</p>
    <p>gender: ${lawyer.gender}</p>
    <p>appointments:</p>
    <ul>
    <g:each var="app" in="${lawyer.appointments}">
         <li>${app.Date}</li>
    </g:each>
    </ul>
</g:each>

请记住，Grails默认使用延迟抓取，因此可能不会填充约会字段，并且需要额外的查询。 See Grails fetching DSL docs

Answer 2

Id是自动生成的字段，其类型为Long，因此您可以从两个模型中删除它。您还可以更新型号：

Lowyer模型：

String firstName
String gender
String school

static hasMany = [appointments: Appointment]

约会模式：

Date appDate
Date appTime

static belongsTo = [lowyer: Lowyer]

然后你可以使用像lower.appointments这样的东西。

<强>更新

<g:each var="appointments" in="${lawyer}">
    <p>
        firstdsdsdsName:
        ${lawyer.firstName}
    </p>
    <p>
        gender:
        ${lawyer.gender}
    </p>
    <p>appointments:</p>
    <ul>
        <g:each var="app" in="${lawyer.appointments}">
            <li>
                ${app.appointmentTime}
            </li>
        </g:each>
    </ul>
</g:each>

现在我认为您的问题是<g:each var="app" in="${lawyer.appointments}">尝试类似<g:each var="app" in="${appointments}">

的问题

组合2个模型并在视图中显示

2 个答案: