douni1396 2018-06-02 14:49
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在PHP中定义动态表单ID

I want to define dynamic form id for the following example form.

The following code doesn't give me the result I want. for example, instead of <form id="form1">,
I get this result : <form id='form''1'>

How should I write it so it gives me the result I want? Thanks !!

<?php 

    while ($data = mysqli_fetch_array($subject)) {

    echo "<form id='form'.'{$data['productID']}' method='post' action=' ' data-checktable='{$data['productID']}'>";
        echo "<input type='checkbox'  class='SelectAll'>All";
        echo "<label><input type='checkbox'  class='selector' value='{$data['product1']}'>" . $data['product1']."</label>";
        echo "<label><input type='checkbox'  class='selector' value='{$data['product2']}'>" . $data['product2']."</label>";
        echo "</form>";
         }
    ?>
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1条回答 默认 最新

  • douling6469 2018-06-02 15:03
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    You just need to Change it to

    echo "<form id='form{$data['productID']}' method='post' action=' ' data-checktable='{$data['productID']}'>";
    

    i.e. remove the extra '.' between the word form and the ID you wanted.

    本回答被题主选为最佳回答 , 对您是否有帮助呢?
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