i have a dropdown box and i already upload the value from dropdown selection to database. after uploading i want to cange the dropdown value how should i do,please help me.

thanks in advance

3 Years
Discussion Span
Last Post by malatamil

this code is to upload values to database table.

<select id="city" name="city" onChange="getPlace(this.value)">
        <option value="" selected="selected">... Select City ...</option>
                $query = "SELECT `state_id`, `state` FROM `states` ";
                $result = mysql_query($query);

    //echo mysql_error();
    while($rows = mysql_fetch_array($result))
    $selected = "";
    $categoryId = $rows['state_id'];
    $categoryName = $rows['state'];
    if($catid == $categoryId)
    $selected = "selected";
            <option value="<?php echo $categoryId; ?>" <?php echo $selected; ?> ><?php echo $categoryName; ?></option>
            <?php } ?>
<div align="left" id="citydiv">

and javascript to display checkbox value

function getPlace(stateId) {        

        var strURL="findplace.php?city="+stateId;
        var req = getXMLHTTP();

        if (req) {

            req.onreadystatechange = function() {
                if (req.readyState == 4) {
                    // only if "OK"
                    if (req.status == 200) {                        
                    } else {
                        alert("There was a problem while using XMLHTTP:\n" + req.statusText);
            req.open("GET", strURL, true);


<? $city=intval($_GET['city']);
$query="SELECT `city_id`, `city` FROM `cities` WHERE `state_id`='$city'";

<td><div align="left">Place:</div></td><td>

<? while($row=mysql_fetch_array($result)) { ?>
<input type="checkbox"  name="place[]" value="<?=$row['city_id']?>" /><?=$row['city']?>

<? } ?>

$lid= mysql_insert_id();                   

            //cityid and placeid added here
            $cityid = trim(ucwords(strtolower(mysql_real_escape_string($_POST['city']))));
            foreach($_POST["place"] as $i=>$value) 
                $sql3 = mysql_query("INSERT INTO `city_place`(`city_id`, `place_id`, `spaid`) VALUES ('$cityid','$value',$lid)");

after uploading in edit page i want to show the selected value in dropdown selection and checkbox selected value only,then if i want to change the selected value i should select from dropdown and checkbox


i have no idea about update values in dropdownbox and checkbox. above i have added insert query to database. so please guide me. i tried but am not getting exactly.where i have to add ajax call function and how to use it for selected cities and also non selected cities in checkbox based on dropdown menu selection. in dropdown menu i want to show only one city and in checkbox to show both selected and non selected cities.

 <?php $sql3 = mysql_query("SELECT s.id,cp.`id`, cp.`state_id`, cp.`city_id`, cp.`spaid`,c.city,st.state FROM spa s 
                LEFT JOIN `city_place` cp ON cp.spaid=s.id
                LEFT JOIN states st ON st.state_id=cp.state_id
                LEFT JOIN cities c ON c.city_id=cp.city_id
                 WHERE s.id='$id' LIMIT 0,1");
                while($c = mysql_fetch_array($sql3))

                <tr><td>City: </td><td><select name="city" id="city" onChange="getPlace(this.value)">
                <option value="" selected="selected"><?php echo $c['state']; ?></option>
            <?php } ?>

            <tr><td><div id="placediv">
                <ul class="subcat_tick">
                    <? while($row = mysql_fetch_array($sql11))
                    $state_id = $row['state_id'];
                    $city_id = $row['city_id'];
                    $checked = (is_null($city_id)) ? '' : " checked";  ?>
                    <input name='place[$state_id]' type='checkbox'$checked /> <span><?=$row['city']?></span>

                    <? } ?>

Edited by malatamil: updation

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