don11

Checking the results after executing the query worked. Thanks alot phpfreaks staff helped me again. Thanks once again.

Hello, I have a problem. I run this query: $query = "SELECT column1 FROM table WHERE column2='$kk'"; $which = $link1; mysql_query($query,$which); some codes.... It is displaying all data correct but only when $kk exists in column2 and when $kk does not exist, it display blank page. So, i want to do a check before running above query to make sure that $kk exists in column2. How to check it before running query?

Thanks alot Muddy_Funster, now i got it. Thanks alot.

I was trying and i have a problem. Please take a look at this code: //doing a query from db1: $query = "SELECT invitation FROM contact"; $which = $link1; mysql_query($query,$which); $result="$kk"; How to store the result in $kk?

Hello, I need help. I want to know how to copy a field value of a table from one database to other database using php. I have a field name "invites" in a table "members" in database "db1" and a field name "new_invites" in a table "new_members" in database "db2". I want to copy value of field "invites" to "new_invites". How to do that?

Thank you for the reply, problem solved.

I was trying to submit with php and it was working but i was not able to show "Thank you" message from above javascript. How to do that?

i was trying to submit values name, mobile, email, msg and send an email to me but i am not able to submit those values from below java code, how to do that. Plz help. <script type="text/javascript"> $(document).ready(function(){ var feed_width = $('#feedback').width(); var scr_w = screen.width; // Screen Width // 26 is width of the veritcal feedback button var btn_width = 26; var move_right = scr_w - btn_width; var move_left = -(feed_width - btn_width); var slide_from_left = 0; var slide_from_right = scr_w - (feed_width - btn_width); var center = ( scr_w / 2 ) - ( feed_width / 2 ); // Positioning the feedback form at the time of page loading positioningForm(); // Handling the right_btn and lift_btn event animations $('.right_btn').click(function(){ slideFromRight(); }); $('.left_btn').click(function(){ slideFromleft(); }); // Moving left or right by clicking close button $('.feed_close').click(function(){ var pos = $('#feedback').position(); var ls = pos.left; if(ls == slide_from_left){ // feedback form is at LEFT moveRight(); }else if(ls == center){ // feedback form is at RIGHT moveRight(); }else{ // feedback form is at CENTER moveLeft(); } }); $('#submit_btn').click(function(){ var name=$('#name').val(); var mobile=$('#mobile').val(); var email=$('#email').val(); var msg=$('#msg').val(); if(name.length>0 && mobile.length>0 && email.length>0 && msg.length>0) { $('.left_btn').hide(); $('.right_btn').hide(); $('.box').hide(); $('#feedback').animate({left: center+"px"},{duration: 'slow',easing: 'easeOutElastic'}); $('.thankyou').show(); } else { $('#error').html('All fields are required!'); } return false(); }); function positioningForm(){ $('.left_btn').hide(); $('#feedback').css({"left": move_right+"px"}).show(); } function slideFromRight(){ $('#feedback').animate({left: slide_from_right+"px"},{duration: 'slow',easing: 'easeOutElastic'}); $('.left_btn').hide(); } function slideFromleft(){ $('#feedback').animate({left: slide_from_left+"px"},{duration: 'slow',easing: 'easeOutElastic'}); $('.right_btn').hide(); } function moveLeft(){ $('#feedback').animate({left: move_left+"px"},{duration: 'slow',easing: 'easeOutElastic'}); $('.left_btn').show(); } function moveRight(){ $('#feedback').animate({left: move_right+"px"},{duration: 'slow',easing: 'easeOutElastic'}); $('.right_btn').show(); } }); </script>

problem is solved now, it was line break issue, thanks

echo shows everything fine and in phpmyadmin, showing same error msg, what to do now

I am getting this error while trying to insert data into table from html form: Could not enter data: You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near 'contact,name,age,email' at line 1 I was trying like this: $sql = "INSERT INTO mytablename ". "(contact,name,age,email,............so on...to 108 colonms) ". "VALUES ". "('$contact','$name','$age','$email',........so on....to 108 values)";

Problem is solved now, thanx brother.

Hello, I am using a javascript to load region, country and city in select box but values are not submitting. This is the code that i am using: From:<br/> Region: <select onchange="set_country(this,country,city_state)" size="1" name="region" id="firegion" class="DEPENDS ON gt BEING inter"> <option value="" selected="selected">SELECT REGION</option> <script type="text/javascript"> setRegions(this); </script> </select><br/> Country: <select name="country" size="1" disabled="disabled" onchange="set_city_state(this,city_state)" id="ficountry" class="DEPENDS ON gt BEING inter"></select><br/> City/State: <select name="city_state" size="1" disabled="disabled" onchange="print_city_state(country,this)" id="ficity" class="DEPENDS ON gt BEING inter"></select> In above code, all data are loaded from external data.js file

Hello, I need a clue to find out the solution of my problem. On selecting the countries, states were loading well before. But when i made some changes to css, states are not loading. I really don't know what is the problem.

How will you code if you have to do that? sample plz