Sign In!
|
New User?Signup Now!
|
Forgot password?
Source Code 4 you
Home
Create Forum Topic
Check WebSite Rank
Publish Articles
UV Web Booster
Contact
Forum Type
Search Engine
.Net C# VB F# VC++
MonoPoject
Java J2EE
Zand Php Cake Php
Oracle Database Server
Mysql Database Server
DB2 Database Server
MSSql Databse Server
Javascript Ajax
Html Xml Xhtml
C Programming
C++ Programming
D Programming
CGI Programming
Perl Programming
Cobol Programming
Pascal Programming
Ruby Programming
Drupal Programming
Automobile
Software Testing
Linux Unix Mac Sun OS
Embedded Systems
Mainframe
IC Microcircuit Chip
Electronics Electrical
Hardware Netwarking
SAP ERP
Logical Reasoning
Human Resources
Science
Agriculture
Accounting
History
Pharmaceutical
Other Topic
Zand Php Cake Php
Posted by :
Iv
Javascript var to php and mysql
I am having some error when i try to pass a variable via javascript
call to a php function.
I am trying to pass a state name to php so that it can search for all
records and pull it back with that name.
I keep getting an error saying:
Error displaying info: Alabama which tells me the value is being
passed but it is not being executing well as the query. Can anyone
assist with the code below and tell me what the problem is.
<script type="text/javascript">
function showlistingsbystate(thestate)
{
var thestate="Alabama";
document.getElementById('
resultsTable').innerHTML = '<?php
displayCarListingsbyState("'+
thestate+'");?>';
}
</script>
<?
/**
* displayCarListings - Displays the car database table in
* a nicely formatted html table.
*/
function displayCarListingsbyState($
thestate){
global $database;
echo $thestate;
$q = "SELECT
carid,zipcode,cartype,year,
carmodel,state,mileage,
timestamp "
."FROM ".TBL_SUBMITTED_CARS." WHERE state = '$thestate'";
$result = $database->query($q);
/* Error occurred, return given name by default */
$num_rows = mysql_numrows($result);
if(!$result || ($num_rows < 0)){
echo "Error displaying info" .$thestate;
return;
}
if($num_rows == 0){
echo "Database table empty";
return;
}
/* Display table contents */
echo '<img src="img/table_top.gif" width="901" height="11" /
><table width="901" cellspacing="0" cellpadding="0">';
echo '<tr class="headerRow"><th width=200>Make</th><th align=right
width=25><a rel="nofollow" href="#">Model</a></th><th align=right
width=20><a rel="nofollow" href="#">Year</a></th><th align=right
width=25><a rel="nofollow" href="#">Mileage</a></th><th align=right
width=25>Zipcode</th><th align=right width=25>State</th><th
align=right width=25><a rel="nofollow" href="#"><img src="img/
arr_desc.gif" border="0" width="16" height="16"> Posted</a></
th></
tr>';
for($i=0; $i<$num_rows; $i++){
$carid = mysql_result($result,$i,"
carid");
$cartype = mysql_result($result,$i,"
cartype");
$carmodel = mysql_result($result,$i,"
carmodel");
$year = mysql_result($result,$i,"year"
);
$state = mysql_result($result,$i,"
state");
$zipcode = mysql_result($result,$i,"
zipcode");
$mileage = mysql_result($result,$i,"
mileage");
$time = mysql_result($result,$i,"
timestamp");
if ( $i&1 ){
echo '<tr onclick="alert(\'Loading Car Page\')" title="'.
$cartype.'" class="evenRow" onmouseout="this.className=\'
evenRow\';"
onmouseover="this.className=\'
rollOverRow\';"><td>'.$
cartype.'</
td><td>'.$carmodel.'</td><td>'
.$year.'</td><td>'.$mileage.'<
/td><td>'.
$zipcode.'</td><td>'.$state.'<
/td><td>'.date('Y-m-d', $time).'</td></
tr>';
}
else{
echo '<tr onclick="alert(\'Loading Car Page\')" title="'.
$cartype.'" class="oddRow" onmouseout="this.className=\'
oddRow\';"
onmouseover="this.className=\'
rollOverRow\';"><td>'.$
cartype.'</
td><td>'.$carmodel.'</td><td>'
.$year.'</td><td>'.$mileage.'<
/td><td>'.
$zipcode.'</td><td>'.$state.'<
/td><td>'.date('Y-m-d', $time).'</td></
tr>';
}
}
echo '</table><img src="img/table_bottom.gif" width="901"
height="11" /><br>';
}
?>
Posted by :
PatiDure
Re : Javascript var to php and mysql
Javascript code is on Your browser and PHP codein the server, so
there's no way that You can call PHP function from JS... Plus they are
totally different languages.
You must Use eg jQuery to send AJAX call from JS code to the server to
reach the PHP function.
Posted by :
Gaurav Kumar
Re : Javascript var to php and mysql
Ohhh.. You can not work directly with Javascript and PHP in the way.
You require a medium for the interaction of Javascript (client Side) and PHP (At server end).
You require Ajax for this purpose
Posted by :
TJ
Re : Javascript var to php and mysql
------------------------------
------------------------------
------------------
To put the cart before the horse, capturing a php variable into
javascript is accomplished by embedding the php tags inside the
javascript tags, as demonstrated in this example (saved and accessed
in a webpage as an *example.php* file):
<?php
$php_variable = "php_string_or_value"; // (php variable may
also come from a URL pass, form POST, or be returned from a php
function.)
?>
<script type="text/javascript">
var javascript_variable_from_php_
variable = "<?php echo
$php_variable ?>";
document.write (javascript_variable_from_php_
variable);
</script>
------------------------------
------------------------------
------------------
As for the horse, the same strategy doesn't work in reverse, though,
since javascript tags aren't functional inside php tags (at least in
my hands). However, there is the capacity to embed javascript tags
within a *form* and, by posting, capture a javascript variable as a
hidden php variable, as in the following example which measures a
user's screen resolution:
<form name="javascript_variable_
capture" action="processing.php"
method="post">
<script type="text/javascript">
var screen_width = screen.width;
var screen_height = screen.height;
document.write ("<input type='hidden' name='screen_width'
value='"+screen_width+"'>");
document.write ("<input type='hidden' name='screen_height'
value='"+screen_height+"'>");
</script>
</form>
and, of course, the submit button:
<input type="submit" name="submit" value="enter" >
and, then, in a *processing.php* file:
$screen_width = $_POST['screen_width'];
$screen_height = $_POST['screen_height'];
This method requires the user to *submit* the form entries to POST.
There are ways to *auto-submit forms* suggested on the Net (e.g. <body
onload="submitForm()">) but I have not worked out the timing to insure
the embeded javascript code is acted upon before the auto-submit
code. Perhaps one needs to start with FORMS constructed with
Javascript -- I have seen reference of such things, but have no
experience. Luck.
Posted by :
TJ
Re : Javascript var to php and mysql
I refined the code of my previous reply:
To pass a javascript variable into php, use a FORM POST with auto-
submit since JavaScript code cannot be embedded into PHP code. Here
is an expample using an a html form with an auto-submit. This method
requires two files, the first called as the webpage to begin the
process, and the second as the answering page to demonstrate that the
javascript variable has be captured by the server's php cache.
For example, name the first file javascript_var_to_php.htm as the
first webpage called, with the following code:
<form ID="variable_pass" name="javascript_variable_to_
php_variable"
action="variable_capture.php" method="post">
<script type="text/javascript">
var javascript_variable = "javascript_string_or_value";
document.write ("<input type='hidden'
name='javascript_variable' value='"+javascript_variable+"
'>");
</script>
</form>
<script type="text/javascript">
document.forms("variable_pass"
).submit()
</script>
Then, the information is passed to the php in the server and can be
demonstrated with the following file -- call it variable_capture.php
-- saved in the same folder as the first, with the following code:
<?php
$php_variable_from_javascript_
variable = $_POST
['javascript_variable'];
function display_captured_js_var_into_
php()
{
global $php_variable_from_javascript_
variable;
echo ("The JavaScript variable 'javascript_variable' passed
to the PHP ")
echo ("variable '\$php_variable_from_
javascript_variable' is
the string '");
echo ($php_variable_from_
javascript_variable);
echo ("'.");
};
display_captured_js_var_into_
php();
?>
==============================
============================
To capture a PHP variable into JavaScript, on the other hand, simply
embed the PHP code into JavaScript tags. As an example, call up the
webpage coded with a file -- call it php_var_to_js.php -- with the
following code:
<?php
$php_variable = "php_string_or_value";
?>
<script type="text/javascript">
var javascript_variable_from_php_
variable = "<?php echo
$php_variable ?>";
document.write ("The PHP variable '$php_variable' passed to the
JavaScript ") +
document.write ("variable 'javascript_variable_from_php_
variable'
is the ") +
document.write ("string '" +
javascript_variable_from_php_
variable + "'.");
</script>
==============================
============================
luck,
TJ
If you have the better reply, then send it to us. We will display your reply after the approval.
Name :
Email Id :
Reply :