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alena1347

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  1. sorry I was not able to explain properly. The remaining dropdows that are after the query are not been loaded though they have nothing to do with php....
  2. I have 3 drop downs and as soon as the code passes from this query it stops and other all dropdowns are loaded on browser and the later part does not have any php code. QUeries till now showed error but this one is not showing a error too
  3. Help me with this simple query please 1)The code below does not show any result and the page dies after that query 2)The below code does not show any error and it just stops the page to load after the query 3)The main thing here is it does not show any error and stops executing what could be the problem. <?php include('conn.php'); $sqls=mysql_query("SELECT * FROM languages")or die(mysql_error()); while($infos=mysql_fetch_array($sqls)) { ?> <option value="<?php echo str_replace(" ","",$infos['name']); ?>"><?php echo $infos['name']; ?></option> <?php } ?>
  4. the data is now refreshed but it took the data and images which are not on my server or database .I tried to use the like from many other pc's but the data fetched is always the old one. please note that the data it took where on my server and database in the past which are now deleted is this due to facebook caching data or something like that.
  5. I have integrated a facebook share button with the script from developers and now when I use the share button the data fetched is the old data which is not even in my database what can be the problem. I cleared all the cache data from browser still it takes the data that has already been deleted from the website. Please help me,I am not getting the problem.
  6. I have a function called after some data and the function is given below:- now the problem here is that at the #vote replace the color of the text is not assigned as it should be from where it is loaded I.e it is not red and not as the font size it should be. sorry this could be simple but not getting it. I have also tried to give .css to vote but it is still not applying that function attach() { var id="<?php echo $_REQUEST['id'];?>"; var state="<?php echo $_REQUEST['state'];?>"; $.ajax({ url: "petition_details.php", data: {id:id,state:state} }).done(function(result){ $("#spprt").replaceWith($(result).find("#spprt").html()); $("#vote").replaceWith($(result).find("#vote").html()); //The problem is here $("input[type=text]").val(""); $("textarea").val(""); if($("#ttt").text()=="You supported it before") { $('#ttt').css({'background-color':'#FFCC00','font-size':'16px','border-style':'none'}); } else if($("#ttt").text()=="Thank you for your VOTE") { $('#ttt').css({'background-color':'#00FF00','font-size':'16px','border-style':'none'}); } }); }
  7. OK thank you got it the buttons work fine now thank you.
  8. thank you was missing something this solved it.
  9. What I am trying to do is. 1. A button named decline is clicked which on click a jquery event is trigerred which puts a .load() which do not return anything. 2. After the load I need to refresh the current page which I do with .load() function. After which all the buttons stop working. below is the code please let me know where to add the .on() so that all buttons work proper. $('#decline').click(function(){ var dec=[]; $("input[name=checkbox]:checked").each(function(){dec.push($(this).val());}); $("#dec").load("list.php",{fors:"decline",arr:dec}) ; $("#td1").load("pending_approvals.php #td1"); }); Thank You!
  10. I used .on but the buttons on that page still do not work.
  11. I am stuck at this code I am using the jquery to load a entire id which works ok without live but for some reason I need to add live, But after using it the whole document buttons fail to work please help with this. $(document).ready(function(){ $("#td3 a").live("click",function(e){ var id=$(e.target).text(); $("#td1").load("pending_approvals.php #td1",{page:id}); }); }); <tr> <td id="td3"> Enter View Limit:<input type="text" name="limit" /> Pages>> <?php for ($i = 1; $i <= $total_pages; $i++) { echo "<a href='#'>$i</a> "; } ?> </td> </tr>
  12. But the above will select just id 1 I need the data of all three
  13. Sorry this could sound simple but not getting it. I need help on writing mysql query $sql=mysql_query("SELECT * FROM TABLE WHERE id = ("some array")"); the above query need to be run with id matches with a array of values,the array is say not of definite size. please could someone help
  14. Thank you this is solved i need to rename the values with another array and it will rename it.
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