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Member Avatar for jithusdani

HI folks.... I want to create a form that includes a textBox and a button,when ever i will click the button i shuld want to create a new text box.and i want to take the value of the each text box when a second button click.If anybody knows the answer …

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Member Avatar for jonasjo

[CODE=php]<?php $server="localhost"; $username="root"; $password="root"; $db_name="test"; $db=mysql_connect($server,$username,$password); mysql_select_db($db_name,$db); if(isset($_POST['btnsave'])) { $quesid=$row['question_id']; $question=$_POST['txtquestion']; $ans=$_POST['choice']; $mark=$_POST['mark']; /*checking no of questions in question table */ $countcheck=mysql_query("select count (*) from question"); echo "no of questions in table are".$countcheck; while($row = mysql_fetch_array($countcheck)) { for(i=0;i<='$countcheck';i++) echo $row[i]; } $ab=mysql_query("SELECT * FROM question "); while($row = mysql_fetch_array($ab)) { …

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Member Avatar for jonasjo

I have error in tis code.any help will b appreciable.tanks in advance. if(isset($_POST['btnsave'])) { for($i=0;$i<count($_POST);$i++) { if(!empty($_POST["hdntxtRow".$i])) { echo $_POST["hdntxtRow".$i]."<br />"; } } $question=$_POST['txtquestion']; $choice1=$_POST['txtchoice1']; $choice2=$_POST['txtchoice2']; $choice3=$_POST['txtchoice3']; $choice4=$_POST['hdntxtrow1']; $choice5=$_POST['hdntxtrow2']; $answer=$_POST['txtanswer']; $a=mysql_query("INSERT INTO question(questions,choice1,choice2,choice3,choice4,choice5,answer) VALUES('$question','$choice1','$choice2','$choice3','$choice4','$choice5','$answer')"); } ?> <!DOCTYPE HTML PUBLIC "-//W3C//DTD HTML 4.0 Transitional//EN"> <HTML> <HEAD> <TITLE> New Document </TITLE> <META …

Member Avatar for ko ko
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