Can't insert values in database MYSQLI what I'm doing wrong?

Question

I'm getting message NOT INSERTED all the time. it look all good when I PUT VALUES by my self in "values ('','','')" but to take it from the fields that I want it doesn't work. Can anybody help where I do it wrong?

So I would like to have in my database after clicking a button something like that:

Item       Clicks
TV LG 55'  0.01
TV LG 55'  0.02
TV LG 55'  0.03
TV LG 55'  0.04

HTML FILE

<!DOCTYPE html>
<html>

<head> </head>

<body>
    <form name="form" method="post" action="clicking.php">
        <span id="TVid">TV LG 55' </span>
        <input type="hidden" name="TVid1" id="TVid1" >

        <hr class="soft"/>

        <div class="control-group">
            <label><span id="TvDiv"> 0.00</span></label>
            <input type="hidden" name="TvDiv1" id="TvDiv1" >
        </div>

        <input type="submit" name="b1" value="Quick Click" id="TvButton" onclick="aa(), bb()">
        </div><br>
    </form>
<script src="counting.js"> </script>
<script 
src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.3/jquery.min.js">
</script>
<script src="extract.js"> </script>
</body>
</html>

PHP FILE ( clicking.php ):

<?php


$con = mysqli_connect("localhost", "root", "") or die("not connected");

$mydb = mysqli_select_db($con, "quickclicks") or die("no db found");


if($con){
    echo "connection good";
} else{
    echo "error in cnnection";
}

if($mydb){   
    echo "connection good";
} else{
    echo "error in cnnection";
}

$item = $_POST ['TVid1'];
$clicks = $_POST ['TvDiv1'];

$sql =  "insert into clicks (id,item,clicks) 
            values ('First','$item','$clicks')";

 if(!mysqli_query($con, $sql)) {
    echo 'not inserted';
 }else {
    echo 'Insterted';
 }

 header("refresh:10; url=new.html");
 ?>

Java script file: counting.js to make count of clicks.

var LGtv = 0.00;
function aa(){
    if (LGtv < 0.05){
        document.getElementById('TvDiv').innerHTML = LGtv +=0.01;
    } else {
        document.getElementById('TvDiv').innerHTML = "You are a WINNER!";
    }
}

Jquery file : extract.js for putting values if TVid and TvDiv in INPUT hidden.

 function bb (){

    var s1 = document.getElementById('TVid').innerHTML;
    document.form.TVid1.value = s1;
    alert ("The value of the hidden field is " + document.form.TVid1.value);  
 // for testing

    var s2 = document.getElementById('TvDiv').innerHTML;
    document.form.TvDiv1.value = s2;
    alert ("The value of the hidden field is " + document.form.TvDiv1.value);  
    // for testing
 }

Answer 1

check id in the table is a primary key. if it's primary key then you can't use 'First' constant for all row
values ('First','$id','$item')";

Can't insert values in database MYSQLI what I'm doing wrong?

Question

1 answers

solution1
0 2017-09-17 06:39:26

Can't insert values in database MYSQLI what I'm doing wrong?

Question

1 answers

solution1 0 2017-09-17 06:39:26

solution1
0 2017-09-17 06:39:26