PHP is executed on the server.
Javascript is executed on the client side (in the browser).
If you want to create real dynamic pages, I think you should learn Javascript as well. Also jQuery (Javascript library) and AJAX.
PHP is executed on the server.
Javascript is executed on the client side (in the browser).
If you want to create real dynamic pages, I think you should learn Javascript as well. Also jQuery (Javascript library) and AJAX.
Check jqGrid if it is what you need.
http://www.trirand.com/jqgridwiki/doku.php
http://trirand.com/blog/jqgrid/jqgrid.html
http://www.trirand.com/blog/?page_id=5
Try instead:
$something = mysql_fetch_array($count);
while ($something = mysql_fetch_array($count)){
write:
while ($something = mysql_fetch_array($result)){
Does it help you?
<select name="table-type" id="select-t">
<option value="" disabled selected>Types of Tables</option>
<option value="seat-r">Round Table</option>
<option value="seat-b">Banquet Table</option>
<option value="seat-s">Square Table</option>
</select><br>
No. of seats : <br>
<input type="text" readonly placeholder="Select Table" size="8" id="inform" />
<select name="seatno" class="seat" id="seat-r" style="display: none;">
<option value="" selected disabled>No. of seats</option>
<option value="6">6</option>
<option value="8">8</option>
<option value="10">10</option>
<option value="12">12</option>
</select>
<select name="seatno" class="seat" id="seat-b" style="display: none;">
<option value="" disabled selected>No. of seats</option>
<option value="6">6</option>
<option value="8"></option>
<option value="10"></option>
</select>
<select name="seatno" class="seat" id="seat-s" style="display: none;">
<option value="" disabled selected></option>
<option value="6">6</option>
<option value="8">8</option>
</select>
<script src="http://ajax.googleapis.com/ajax/libs/jquery/1/jquery.min.js"></script>
<script type="text/javascript">
$(document).ready(function(){
$('#select-t').unbind().bind('change',function(){
var elemId = $(this).val();
showElement(elemId);
$("#inform").hide();
});
function showElement(id){
$('.seat').each(function(){
if($(this).attr('id') == id){
$(this).show();
} else {
$(this).hide();
}
});
}
});
</script>
This should work:
<?php
$i = 0;
while($row = mysql_fetch_assoc($query)){
$dataLevel[$i]['q_id'] = $row['q_id'];
$dataLevel[$i]['response_value'] = $row['response_value'];
$dataLevel[$i]['cr_chpt'] = $row['cr_chpt'];
$dataLevel[$i]['ql_level'] = $row['ql_level'];
$i++;
}
?>
<script type="text/javascript">
var dataLevel = JSON.parse('<?php echo json_encode($dataLevel); ?>');
for(var i=0; i<dataLevel.length; i++){
document.write(dataLevel[i]['q_id'] + '<br />');
document.write(dataLevel[i]['response_value'] + '<br />');
document.write(dataLevel[i]['cr_chpt'] + '<br />');
document.write(dataLevel[i]['ql_level'] + '<br /><br />');
}
</script>
Can this help you?
<?php
$dataLevel = [[280, "ANALYZE", 260, 560],
[230, "APPLY", 260, 510],
[180, "UNDERSTAND", 260, 460],
[130, "REMEMBER", 260, 410]];
?>
<script type="text/javascript">
var dataLevel = JSON.parse('<?php echo json_encode($dataLevel); ?>');
for(var i=0; i<dataLevel.length; i++){
document.write(dataLevel[i] + '<br />');
}
</script>
jQuery('.date-pick').removeClass('hasDatepicker').datepicker({
dateFormat: 'mm-dd-yy'
});
Is the server on Linux? I tested the code on the server that runs on Linux and I had the same problem and after runing in Terminal 'chmod 777 -R mytestsite' it worked fine. I just gave all privileges to the folder that contains the php script file and the image.
Check the picture properties, it may be read-only.
Then you may change the code to:
$old = "./images/general/item2.jpg";
$new = "./images/general/item44.jpg";
if(file_exists($old))
{
rename($old, $new) or die('Error renaming file.');
echo "The file has been renamed successfully.";
}
else
{
echo 'The file does not exist.';
}