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NebuJohn

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  1. Somebody please help me. The below given code is not working. $(document).on('keyup', ".add_commi", function() { typingTimer = setTimeout(doneTyping, doneTypingInterval); var pro = $(this).attr('id'); var selectvalue = $(this).html(); var id = pro.slice(-2); var amount = $('#amount'+id).html(); var rate = $('#v_rate'+id).html(); //alert('working'); var data = 'cur_amount='+cur_amount +'&cur_rate='+cur_rate+'&commision='+selectvalue; $.ajax({ type: "POST", dataType: "json", url: "http://localhost/agent/buy-currency-calculations/", data: data, success: function(output){ var cus_rate = output.cus_rate; var total = output.total; var tot_commision = output.commision; alert(cus_rate, total, tot_commision); $('#curr_rate'+id).html(cus_rate); $('#tot'+id).html(total); $('#tot_commi'+id).html(tot_commision); } }); return false; }); The problem is with Ajax calling. Can somebody give me a hint to rectify this. Thanks in advance.
  2. Thanks for the reply DrTrans.. @JaysonDotPh Here goes my solution <input type="radio" name="stall" value="41" <?php require ("database.php"); $set = 0; $result = mysqli_query($con,"SELECT * FROM tab1"); $row = mysqli_fetch_array($result); $data = $row['sadlle']; $data = unserialize($data); $max = sizeof($data); for($i=0; $i<$max; $i++){ if($data[$i]==41) { $set = 1;} } if($set==1) { echo "disabled";} ?>>
  3. Hi, I have a set of radio buttons with some integer as its value. I use them to make seat selections in an application. If a Radio button is once selected its value will be entered to the DataBase. My question, can I mask or make invisible, the radio button whose value is found in database? isa that possible? Any help will be appreciable. Thanks in advance.. Regards...
  4. Hi, I dont know if this is a wrong category to post. I have a registration form in my website. I need a overlay popup after the sucessfull registration process. Can someone help me. Thanks in advance. Regards..
  5. Such an option is not present there...
  6. Can someone help me to set my own signature in phpfreaks. Thanks in advance.
  7. Hi, I need a PHP uploader. I need the uploader to upload the file simultaneously when the images is browsed and submitter. The html form will be submitted after that. Can some one help me? any help will be appriciable. Thanks in advance.
  8. I have recived a set of dat from an android device using Json. I have decoded this data into a PHP array. I took each value of the array usinf foreach and placed in a MySQL statement. But its not working. Can any body help me. Given below is my code. $array=$_POST['nuer']; //numbers fetched $jstring = (array)json_decode($array); foreach($jstring as $value ) //loop over values { $four = mb_substr($value, 0,-10); //to get the first chara's $ten = mb_substr($value, -10); // to get the last ten chara $con=mysqli_connect("localhost","app","test","appt"); //connection to the database $result = mysqli_query($con, "SELECT * FROM profile WHERE (mobileno IN '$ten') AND (code IN '$four')"); while($row = mysqli_fetch_array($result, MYSQL_ASSOC)) { $friend[] = $row[id]; } mysqli_close($con); $result is returning a null value. When i used a user defined array with random value, this code worked perfectly. Thanks in Advance.
  9. I had to implode and explode because it have to be comma seperated. I have to retrive each contacts from mobile contacts from an android device.
  10. I have retrived a set of numbers from a device, I divided it into two parts and compared if that exist in a table. When i gave a random array it worked properly. Below I give the code. <?php $array=$_POST['tel']; //numbers fetched from contact of mobile $jstring = (array)json_decode($array,JSON_FORCE_OBJECT); $ttutt =implode("','", $jstring); $tet = explode(',', $ttutt); foreach( $tet as $value ) //loop over values { $four = mb_substr($value, 0,-10); //to get the first chara's $ten = mb_substr($value, -10); // to get the last ten chara $corn=mysqli_connect("localhost","bukado","********","frocd"); //connection to the database $result = mysqli_query($corn, "SELECT * From prip WHERE nob IN ($ten) AND cdac IN ($four)"); while($row = mysqli_fetch_array($result, MYSQL_ASSOC)) { $forl[] = $row[no]; } mysqli_close($corn); } ?> Can any body please help me? any help will be appreciable. Regards. Thanks in advance.
  11. I fixed this problem. Sharing with others. $array=$_POST[number]; $jstring = (array)json_decode($array,true); $sa = "'".implode(',',$jstring)."'";
  12. May i Know what all information you need more. number is a Jason object which holds numbers from mobile contact. I was able to append that into a file. Now i need that data to be fetched into a SQL query as comma separated string.
  13. Below Given is my Code. $array=$_POST[number]; $jstring = json_decode($array,true); $sa = "'".implode("','",$jstring)."'"; The error is given below. Warning: implode() [function.implode]: Invalid arguments passed <root> on line <line no> 'number' is a JSON object.
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