JavaScript

Re: Ajax request call using jquery

21-Mar-13 8:33

Haven't you spotted the spelling mistake?

hdnNextPervious *Previous*

"It's true that hard work never killed anyone. But I figure, why take the chance." - Ronald Reagan

That's what machines are for.

Got a problem?
Sleep on it.

jkirkerx21-Mar-13 9:24

jkirkerx

21-Mar-13 9:24

I just spent another 5 minutes looking for it, and just caught it.

nice job dusty!

[edit]

Oh, I copied the mistake into my code sample, even worse!, oh well

Slide show

Tichaona J21-Mar-13 0:10

Tichaona J

21-Mar-13 0:10

I’m trying to get my website to show images one after the other using java script I got from www.javascriptkit.com/howto/show2.shtml and have followed there example as closely as I can but with no luck.

HTML

<head>
 <meta http-equiv="content-type" content="text/html; charset=utf-8" />
 <title>Acrele - Home</title>
 <meta name="keywords" content="" />
<meta name="description" content="" />
<link href="CSS1.css" rel="stylesheet" type="text/css" />

<script type="text/javascript">

        var image1=new Image()
        image1.src="16.jpg"

        var image2=new Image()
        image2.src="employee-training.jpg"

        var image3=new Image()
        image3.src="Real_Estate_Investment_Education.jpg"

        var image4=new Image()
        image4.src="Unti2tled.jpg"

        var image5=new Image()
        image5.src="Untitled.jpg"

        var image6=new Image()
        image6.src="zclassroom.jpg"

</script>

</head>

2. Added the html code were I want the image to show:

HTML

<div id="left-feature-colum">

                     <div id="slider-bg">

                        <div id="slider">


                                    <img src="16.jpg" id="slide" width="600px" height="300px">

                         </div>

               </div>

            </div>

3. Edited there javascript accordingly to the number of images I’m using:

JavaScript

<div id="left-feature-colum">

                     <div id="slider-bg">

                        <div id="slider">

                                    <img src="16.jpg" id="slide" width="600px" height="300px">

                                    <script>
                                    var step=1;

                                  //if browser does not support the image object, exit.
                                  var img = document.getElementById('slide');
                                  img.src = "image" + step + ".src";
                                  console.log(img.src);
                                  if (step < 6) step++;
                                      else step = 1;
                                      //call function "slideit()" every 2.5 seconds
                                      setTimeout(slideit, 3500);


                             </script>
                         </div>

               </div>

            </div>

This isn’t working, it only load the first image “16.jpg” only…Help

Re: Slide show

Graham Breach21-Mar-13 1:37

Graham Breach

21-Mar-13 1:37

You have removed the code from inside the "slideit" function, so now there is no way for it to repeat and the setTimeout(slideit, 3500) is trying to call a function that doesn't exist.

Re: Slide show

jkirkerx21-Mar-13 7:55

jkirkerx

21-Mar-13 7:55

There are a lot of bad slideshow scripts out there, I rewrote one that was a CMS plugin for word press which had many errors in it, and made no sense to me. It might be too advanced for you, and requires JQuery, but is simplistic in design.

[Display Advertising Filmstrip^]

Re: Slide show

alisener198022-Mar-13 9:41

alisener1980

22-Mar-13 9:41

you must use setinterval...

A question to dust_dex

Hakan Bulut20-Mar-13 22:51

20-Mar-13 22:51

Your profile is not active so I'm can not know your name. I have a question ..

dusty_dex21-Mar-13 5:29

21-Mar-13 5:29

I have my account back now.

Also changed my profile to allow email replies.

Smile | :)

"It's true that hard work never killed anyone. But I figure, why take the chance." - Ronald Reagan

That's what machines are for.

Got a problem?
Sleep on it.

Hakan Bulut24-Mar-13 21:13

24-Mar-13 21:13

XML

<html>
<head>
<script type="text/javascript">
    var rec = ["19","20","21","40","41","45","12","23","34","36","37","46"]
    document.write("Weekly records for 2 week : ");
    for(var i=0; i<12; i++) {
        document.write(rec[i] + ",");
    }
    document.write("<BR>");
    for(var z=0; z<6; z++) {
        var q = Math.round(12 * Math.random());
        document.write(rec[q], + "," + z);
    }
</script>
</head>
<body>
</body>
</html>

I wrote a simple random permutations, but May Have NaN Error Can help me with this?

modified 25-Mar-13 3:23am.

dusty_dex24-Mar-13 21:53

24-Mar-13 21:53

Off by one error - range of random 0-12, valid range for Array 0 - 11

Removed an unwanted comma (,) after rec[q] in your document write. Smile | :)

<html>
<head>
<script type="text/javascript">
    var rec = ["19","20","21","40","41","45","12","23","34","36","37","46"];
    document.write("Weekly records for 2 week : ");
    for(var i=0; i<12; i++) {
        document.write(rec[i] + ",");
    }
    document.write("<BR>");
    for(var z=0; z<6; z++) {
        var q = Math.round(11 * Math.random());
        document.write(rec[q] +", "+ z +" ");
    }
</script>
</head>
<body>
</body>
</html>

"It's true that hard work never killed anyone. But I figure, why take the chance." - Ronald Reagan

That's what machines are for.

Got a problem?
Sleep on it.

Hakan Bulut24-Mar-13 23:47

24-Mar-13 23:47

Well.. now how do i prevent adding the numbers of array elements value, array elements only among themselves, I just wanted to combine random..

Output is:
19, 0 23, 1 20, 2 19, 3 34, 4 45, 5

it must be:
19, 23, 20, 19, 34, 45

Why does it wrote the numbers of random array and how do i prevent the same number selection..¿

Quickly remove the shameless

Hakan Bulut25-Mar-13 1:11

25-Mar-13 1:11

I have to special thanks for all ur helps.

dusty_dex25-Mar-13 7:25

25-Mar-13 7:25

Well you aren't putting the (value of) the z variable into the array. Only showing its value.

remove the

+ z +" "

from document.write

"It's true that hard work never killed anyone. But I figure, why take the chance." - Ronald Reagan

That's what machines are for.

Got a problem?
Sleep on it.

Hakan Bulut25-Mar-13 3:21

25-Mar-13 3:21

I tried the following lines at least.. have you got an idea about this statement?

document.write("<BR><BR>" + "Combination is  :" + "<BR>");
do(q) { z += 1;
    var q = Math.round(Math.random()* 11);
            if(q==q) {return q;}
    document.write(rec[q] +", "+ z);
} while(z==6);

Hakan Bulut25-Mar-13 3:22

25-Mar-13 3:22

Is there something to way some simple code to do this?

dusty_dex25-Mar-13 8:53

25-Mar-13 8:53

replace with this.

JavaScript

<script>
    // create and initialise - array to keep track of any number taken
    var taken = new Array(49);
    for (var i=49; i>0; i--) taken[i]=false;

    var lucky = [];
    var rec = ["19","20","21","40","41","45","12","23","34","36","37","46"];
    document.write("Weekly records for 2 week : ");
    for(var i=0; i<12; i++) {
        document.write(rec[i] + ",");
    }

    document.write("<br><hr>");

    var z=0;
    do {
        var q = Math.round(11 * Math.random());
        if(!taken[ rec[q] ]) {
            lucky[z] = rec[q];
            taken[ rec[q] ] = true;
            document.write(lucky[z] + "<br>");
            z++;       
        }
    } while (z < 6);
</script>

this line:

if(!taken[ rec[q] ])

checks if NOT taken (! = not)

"It's true that hard work never killed anyone. But I figure, why take the chance." - Ronald Reagan

That's what machines are for.

Got a problem?
Sleep on it.

Hakan Bulut25-Mar-13 21:21

25-Mar-13 21:21

Excellent code thanks a lot.

dusty_dex25-Mar-13 11:58

25-Mar-13 11:58

Just letting you know what your errors here are. See my other post for new code.

do(q)

can't use expressions at the top of do loop *(q)*, just use

do {

z is not declared with var *and* initialised with a value. var z = 0;

z +=1; // z (undefined) +1 is still undefined

you don't need the { } braces for a single statement..

if(q==q) return q;

but, q==q just evaluates to true so you might as well have just written this

return q; // if (q==q) {return q;}

I am not able to reply very quickly. On a training course for another 10 days. Be patient.

Please keep trying to write new code as it is easier for me to guess what you are doing. Smile | :)

"It's true that hard work never killed anyone. But I figure, why take the chance." - Ronald Reagan

That's what machines are for.

Got a problem?
Sleep on it.

modified 25-Mar-13 18:05pm.

Hakan Bulut26-Mar-13 21:58

26-Mar-13 21:58

Finally, ask you for anything more in a array of 12 elemented to 49 membered Argument want to match, you can also help me in this topic? JavaScript writing is so hard really. Thank you for your help.
I tried the following lines:

var arrNumbers = [49];
var arrOutside = [];
var arrInside  = ["19","20","21","40","41","45","12","23","34","36","37","46"];
var z=0;
for(var i=0; i<12; i++) { arrInside[i]=false; }  

for(var j=49; j>0; --j) 
{ 
   arrNumbers[j]=j; 

   if(arrNumbers[j] != arrInside[i]) {
      document.write(arrNumbers[j]=arrOutside[z], +  ", ");
      arrInside[i] = true;
      z++;
   }
}

if this is what special things I did not want to deal with it..
I do not want to discuss whom it is could not be shy

dusty_dex27-Mar-13 8:14

27-Mar-13 8:14

Please explain what inside & outside mean.
inside what?
outside what?

arrNumbers = [49]; // array length = 1 ???

I'm guessing this is what you intended
arrNumbers = new Array(49);

Also, you can't assign (=) inside of document.write()

JavaScript

arrOutside[j]=j

JavaScript

arrNumbers[j] = arrOutside[z]

did you mean to show arrNumbers and arrOutside, or copy from one to another?

this does not make sense.

JavaScript

arrNumbers[j] && arrInside[i] = true;

What other programming language do you know?
Because, it might be easier if you write your code in a computer language that you are familiar with.
I will be able to figure it out.

"It's true that hard work never killed anyone. But I figure, why take the chance." - Ronald Reagan

That's what machines are for.

Got a problem?
Sleep on it.

Hakan Bulut27-Mar-13 21:02