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

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

4

1 に答える 1

0

このループは純粋にphpで実行できますが、なぜjavascriptを使用するのですか

<select name="years" >

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

?>

</select>

複雑なプログラミングではなく、物事を簡単にして、他の人が簡単に理解してあなたを良い名前で呼ぶことができるようにします

于 2013-08-04T15:48:26.190 に答える