有条件地反应呈现

现在我得到了它的工作,但我不喜欢的解决办法。我的解决方案是基于这样一个知识,即只有当呈现函数内部的状态发生变化时,组件才会重新呈现(在本例中为this.state.errorMessages)。所以我必须像这样显式地将if条件放在render函数中

renderError() {
    var errArr = [];
    for (var key in this.state.errorMessages) {
        errArr = [...errArr, ...this.state.errorMessages[key]];
    }
    return (
        <div className="alert alert-danger">
            {errArr.map((err) => {
                return <p>{err}</p>
            })}
        </div>
    )
}

renderForm() {
    return (
        <form onSubmit={this.handleRegisterFormSubmit}>
            <div className="form-group">
                <label>First Name</label>
                <input type="text" className="form-control" name="name" placeholder="Name" value={this.state.firstName} onChange={this.handleFirstNameChange} required/>
            </div>
            <div className="form-group">
                <label>Last Name</label>
                <input type="text" className="form-control" name="lastName" placeholder="Last Name" value={this.state.lastName} onChange={this.handleLastNameChange} required/>
            </div>
            <div className="form-group">
                <label>Email address</label>
                <input type="email" className="form-control" id="exampleInputEmail1" aria-describedby="emailHelp" placeholder="Enter email" value={this.state.email} onChange={this.handleEmailChange} />
                <small id="emailHelp" className="form-text text-muted">We'll never share your email with anyone else.</small>
            </div>
            <div className="form-group">
                <label>Password</label>
                <input type="password" className="form-control" name="password" placeholder="Password" value={this.state.password} onChange={this.handlePasswordChange}/>
            </div>
            <div className="form-group">
                <label>Password Confirmation</label>
                <input type="password" className="form-control" name="password_confirmation" placeholder="Password Confirmation" value={this.state.passwordConfirmation} onChange={this.handlePasswordConfirmationChange}/>
            </div>
            <div>
                <button type="submit" className="btn btn-primary">Submit</button>
                <button type="button" className="btn btn-danger" onClick={this.handleCancelClick}>Cancel</button>
            </div>
        </form>
    )
}

render () {
    if (!this.state.errorMessages) {
        return (
            <div>
                {this.renderForm()}
            </div>
        )
    } else {
        return (
            <div>
                {this.renderForm()}
                {this.renderError()}
            </div>
        )
    }
}

我真的不喜欢这种方法,因为如果我有更多的条件重新渲染,这可能会变得很糟糕。我希望有一个解决方案,在实际的渲染函数中没有太多的逻辑,并将其提取出来。例如。。。

renderError() {
    if (!this.state.errorMessages) {
        return;
    }
    var errArr = [];
    for (var key in this.state.errorMessages) {
        errArr = [...errArr, ...this.state.errorMessages[key]];
    }
    return (
        <div className="alert alert-danger">
            {errArr.map((err) => {
                return <p>{err}</p>
            })}
        </div>
    )
}

render () {
    <form onSubmit={this.handleRegisterFormSubmit}>
        <div className="form-group">
            <label>First Name</label>
            <input type="text" className="form-control" name="name" placeholder="Name" value={this.state.firstName} onChange={this.handleFirstNameChange} required/>
        </div>
        <div className="form-group">
            <label>Last Name</label>
            <input type="text" className="form-control" name="lastName" placeholder="Last Name" value={this.state.lastName} onChange={this.handleLastNameChange} required/>
        </div>
        <div className="form-group">
            <label>Email address</label>
            <input type="email" className="form-control" id="exampleInputEmail1" aria-describedby="emailHelp" placeholder="Enter email" value={this.state.email} onChange={this.handleEmailChange} />
            <small id="emailHelp" className="form-text text-muted">We'll never share your email with anyone else.</small>
        </div>
        <div className="form-group">
            <label>Password</label>
            <input type="password" className="form-control" name="password" placeholder="Password" value={this.state.password} onChange={this.handlePasswordChange}/>
        </div>
        <div className="form-group">
            <label>Password Confirmation</label>
            <input type="password" className="form-control" name="password_confirmation" placeholder="Password Confirmation" value={this.state.passwordConfirmation} onChange={this.handlePasswordConfirmationChange}/>
        </div>
        <div>
            <button type="submit" className="btn btn-primary">Submit</button>
            <button type="button" className="btn btn-danger" onClick={this.handleCancelClick}>Cancel</button>
        </div>
        {this.renderError}
    </form>
}

这会引发一个错误,因为它抱怨This.renderError应该作为函数调用。但当我把它当作 this.renderError() 错误永远不会呈现,因为当错误返回时不会自动调用它。

或者,为什么我不能像下面这样做

render () {
    <div>
        <form onSubmit={this.handleRegisterFormSubmit}>
            ...
        </form>
        {if (this.state.errorMessages) {this.renderError()}}
    </div>
}

这会引发控制台错误

未捕获错误:模块生成失败:语法错误:意外标记 (117:13)

本质上,我正在寻找一种解决方案,在render函数中,当状态发生变化时,我可以轻松地显示整个代码块。在Vue,我可以做

<form>
    <input type="text">
</form>
<div v-if="hasError">
    <div class="alert alert-danger">{{something}}</div>
</div>

Can I do something as easy as this in React?