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Remembering options in a select box array after submitting through php

开发者 https://www.devze.com 2023-03-15 01:35 出处:网络
<form method=\"post\"> <select name=\"box[]\"> <option value=\"1\" <?php if ($_POST[\'box[0]\'] == \"1\") echo \"selected=\\\"selected\\\";\"?>>1</option>
<form method="post">
    <select name="box[]">
        <option value="1" <?php if ($_POST['box[0]'] == "1") echo "selected=\"selected\";"?>>1</option>
        <option value="2" <?php if ($_POST['box[0]'] == "2") echo "selected=\"selected\";"?>>2</option>
        <option value="3" <?php if ($_POST['box[0]'] == "3") echo "selected=\"selected\";"?>>3</option>
    </select>
    <p>
    <select name="box[]">
        <option value="1" <?php if ($_POST['box[1]'] == "1") echo "selected=\"selected\";"?>>1</option>
        <option value="2" <?php if ($_POST['box[1]'] == "2") echo "selected=\"selected\";"?>>2</option>
        <option value="3" <?php if ($_POST['box[1]'] == "3") echo "selected=\"selected\";"?>>3</option>
    </select>
    <p>
    <input type="submit" value="Submit">
</form>

When I use box names "box1" and "box2", it works without a problem. What am I doing wrong?

****** EDIT ********

Thanks a lot for your comments, but I actually found the solution myself, even if it doesn't make much sense. Instead of using $_POST['box'][0] and [1] at the if statement, I simply used $box[0] and [1]. Even though it's posted, apparently php sees it as a normal array, and not as some kind of $_POST-array! Working code:

<form method="post">
    <select name="box[]">
        <option value="1" <?php if ($box[0] == "1") echo "selected='selected'";?>>1</option>
        <option value="2" <?php if ($box[0] == "2") echo "selected='selected'";?>>2</option>
        <option value="3" <?php if ($b开发者_开发技巧ox[0] == "3") echo "selected='selected'";?>>3</option>
    </select>
    <p>
    <select name="box[]">
        <option value="1" <?php if ($box[1] == "1") echo "selected='selected'";?>>1</option>
        <option value="2" <?php if ($box[1] == "2") echo "selected='selected'";?>>2</option>
        <option value="3" <?php if ($box[1] == "3") echo "selected='selected'";?>>3</option>
    </select>
    <p>
    <input type="submit" value="Submit">
</form>


Try using $_POST['box'][0] and $_POST['box'][1] ;) Since they are posted as an array, you need to approach them as a array.


Both, elements have same name. Thats the problem. $_POST['box[0]'] , $_POST['box[1]'] , contains the array of the two elements, not the value it self.


Thanks a lot for your comments, but I actually found the solution myself, even if it doesn't make much sense. Instead of using $_POST['box'][0] and [1] at the if statement, I simply used $box[0] and [1]. Even though it's posted, apparently php sees it as a normal array, and not as some kind of $_POST-array! Working code:

<form method="post">
    <select name="box[]">
        <option value="1" <?php if ($box[0] == "1") echo "selected='selected'";?>>1</option>
        <option value="2" <?php if ($box[0] == "2") echo "selected='selected'";?>>2</option>
        <option value="3" <?php if ($box[0] == "3") echo "selected='selected'";?>>3</option>
    </select>
    <p>
    <select name="box[]">
        <option value="1" <?php if ($box[1] == "1") echo "selected='selected'";?>>1</option>
        <option value="2" <?php if ($box[1] == "2") echo "selected='selected'";?>>2</option>
        <option value="3" <?php if ($box[1] == "3") echo "selected='selected'";?>>3</option>
    </select>
    <p>
    <input type="submit" value="Submit">
</form>
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