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Php 获取symfony2控制器中选择标记的所选选项的id_Php_Symfony - Fatal编程技术网

Php 获取symfony2控制器中选择标记的所选选项的id

Php 获取symfony2控制器中选择标记的所选选项的id,php,symfony,Php,Symfony,我有一个只有一个select标签的表单。当用户提交表单时,我希望在控制器中获取所选选项标记的id。我的表格是 <div class="one_third"> <form name="portfolios" action="{{ path('MunichInnovationGroupBundle_portfolio') }}" method="post" > <h2>Select your portfolio</h

我有一个只有一个select标签的表单。当用户提交表单时,我希望在控制器中获取所选选项标记的id。我的表格是

 <div class="one_third">
        <form name="portfolios" action="{{ path('MunichInnovationGroupBundle_portfolio') }}" method="post" >
            <h2>Select your portfolio</h2>
            <p>Change the portfolio to be managed:<br />
                <select name="portfolio" style="width: 265px; height:28px;">
                    <option selected="selected" value="default">Switch Your Portfolio</option>
                    {% for portfolio in portfolios %}
                        <option id={{ portfolio.id }}>{{ portfolio.portfolioName }}</option>
                        {% endfor %}
                </select>
            </p>
            <p>
                <input style="margin-left:100px; height:26px; cursor:pointer; background-color: #2D9AD3 !important; border: 1px solid #067EBD; color: #FFFFFF; text-shadow: 0 -1px 0 #067EBD;" type="submit"class="button2 tooltip" value="Switch">
            </p> 
        </form>
    </div>
var_dump($data)给了我类似的东西

array
 'portfolio' => string 'Umair Portfolio 1' (length=17)

Array
我怎样才能拿到身份证

有什么建议吗


提前感谢

您必须将属性id更改为值

<option value={{ portfolio.id }}>{{ portfolio.portfolioName }}</option>
{{portfolio.portfolioName}

您是否尝试了上述代码?这正是我的建议,选项标记必须具有value而不是id作为属性。如果你想在中发送id的值,你需要在value属性中设置id。rry我很快就发表了评论。。。这对我很有用,谢谢:)
<option value={{ portfolio.id }}>{{ portfolio.portfolioName }}</option>