小编and*_*sem的帖子

我有这个组件.我想向我创建的每个listElement传递一个调用处理程序.如果我像下面这样做bind(this),它可以正常工作.问题是我在控制台中从React收到此警告:bind(): You are binding a component method to the component. React does this for you automatically in a high-performance way, so you can safely remove this call.

var MyList = React.createClass({
  render: function () {
    var listElements = this.props.myListValues.map(function (val) {
      return (
        <ListElement onCallHandler={this.props.parentsCallHandler} val={val} />
        );
    }.bind(this));
    return (
      <ul>
          {listElements}
      </ul>
      );
  }
});

如果我不绑定它,我的孩子们不知道调用处理程序.我能做些什么不同的事情？

PS:

我知道解构分配,比如解释http://facebook.github.io/react/docs/transferring-props.html#transferring-with-...-in-jsx,但我不想使用Harmony.

我有以下React组件,我想在我的select-block中选择一个带有TestUtils的selectElements.我怎么做？

var selectElements = ["type_a", "type_b"];

var SelectElement = React.createClass({
  render: function() {
    return (
      <option value={this.props.type}>{this.props.type}</option>
      );
  }
});

var MyForm = React.createClass({
  handleSubmit: function(e) {
    console.log("submitted");
  },
  render: function () {
    var selectElements = this.props.availableTypes.map(function (type) {
      return (
        <SelectElement key={type} type={type} />
        );
    });
    return (
      <form role="form" onSubmit={this.handleSubmit}>
        <select ref="type" name="type">
          <option value="">-- Choose --</option>
          {selectElements}
        </select>
        <input type="submit" value="Search"/>
      </form>
      );
  }
});

我已经尝试过这样做了:

var myFormComponent = TestUtils.renderIntoDocument(<MyForm selectElements={selectElements} />);
var form = TestUtils.findRenderedDOMComponentWithTag(myFormComponent, 'form'); …

我正在制作一个表格,我需要收音机输入.如何在onSubmit函数中获取已检查的无线电输入,正确的方法是什么？

这是我的代码,我myRadioInput-variable在提交时包含选项A或选项B:

React.createClass({
    handleSubmit: function() {
        e.preventDefault();
        var myTextInput = this.refs.myTextInput.getDOMNode().value.trim();
        var myRadioInput = "How ?";
    },
    render: function() {
        return (
            <form onSubmit={this.handleSubmit}>
                <input type="text" ref="myTextInput" />
                <label>
                    <span>Option A</span>
                    <input type="radio" name="myRadioInput" value="Option A"/>
                </label>
                <label>
                    <span>Option B</span>
                    <input type="radio" name="myRadioInput" value="Option B"/>
                </label>
                <input type="submit" value="Submit this"/>
            </form>
        )
    }
});

我有dropwizard-application(0.7.0),我想运行集成测试.

我使用DropwizardAppRule设置了集成测试,如下所示:

@ClassRule
public static final DropwizardAppRule<MyAppConfiguration> RULE =
        new DropwizardAppRule<MyAppConfiguration>(
                MyApplication.class, Resources.getResource("testconfiguration.yml").getPath());

当我尝试使用它运行以下测试时,它不起作用,因为我没有运行我的迁移.运行迁移的最佳方法是什么？

测试:

@Test
public void fooTest() {
    Client client = new Client();
    String root = String.format("http://localhost:%d/", RULE.getLocalPort());
    URI uri = UriBuilder.fromUri(root).path("/users").build();
    client.resource(uri).accept(MediaType.APPLICATION_JSON).type(MediaType.APPLICATION_JSON).post(User.class, new LoginUserDTO("email@email.com", "password"));
}

组态:

 public class MyAppConfiguration extends Configuration {
@Valid
@NotNull
private DataSourceFactory database = new DataSourceFactory();

@JsonProperty("database")
public DataSourceFactory getDataSourceFactory() {
    return database;
}

@JsonProperty("database")
public void setDataSourceFactory(DataSourceFactory dataSourceFactory) {
    this.database = dataSourceFactory;
}

}