2013-08-04 15:38
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I am so new in html, php, javascript, mysql. i created a drop down menu using java script that contains a list of 2011 to 2510. During update i can not display the stored value that is generated using java script. I searched but found nothing. A portion of code.....

<select name="eca">
            <option id="eca" required="required" class="ecadetail" value="NSS" <?php if ($eca == 'NSS') echo 'selected="selected"';?>">NSS</option>
            <option id="eca" required="required" class="ecadetail" value="NCC" <?php if ($eca == 'NCC') echo 'selected="selected"';?>">NCC</option>
            <option id="eca" required="required" class="ecadetail" value="Cultural" <?php if ($eca == 'Cultural') echo 'selected="selected"';?>">Cultural</option>
        <div class="label">Year</div>
        <div class="inputyear">
            <select name="years" >
                <script language="JavaScript">
                 // loop to create the list
                     var year = 2010
                     for (var i=1; i <=500; i++)
                        document.write("<option>" + year + "</option>");
                    // end JS code hide -->
                    <option  value="<?php if ($year == 'year') echo 'selected="selected"';?>"></script>
    </div> <!-- end of 7th row -->  

the value for eca is working fine. The value year is stored in $year. help please...

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1条回答 默认 最新

  • doumu8911
    doumu8911 2013-08-04 15:48

    This loop can purely done in php then why involving the javascript

    <select name="years" >
     $matchyear = 2010;
    for ($i=2010; $i <=2510; $i++)
    { ?>
    <option  value="<?php echo $i;?>" <?php if ($i== $matchyear) echo 'selected="selected"';?>><?php echo $i;?> </option>   
    <?php }

    Make things easier for you not a complex programming so that other person can understand easily and call you by good names

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