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Make sure $('.middle_container').find('.start_price').val() and $('.middle_container').find('.end_price').val() are returning proper values. Also to set the value of the slider you have to use the same syntax which you are using for setting min/max values. Also

$.get("ajax.php",options,function(data){
    $('.middle_container').html(data);          

    $('#slider-range').slider( "option", "min", $('.middle_container').find('.start_price').val() );
    $('#slider-range').slider( "option", "max", $('.middle_container').find('.end_price').val() );
    $('#slider-range').slider( "option", "value", $('#slider-range').slider("value"));

});

how do i access each of the values separately? because this is a range so there are two handles

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Destroy the slider first, which removes the slider functionality completely. This will return the element back to its pre-init state.

$("#selector").slider("destroy");

After that you can add new values to the slider as,

$("#selector").slider({
    range: "max",
    min: 0, // min value
    max: 200, // max value
    step: 0.1,
    value: 200, // default value of slider
    slide: function(event, ui) {
        $("#amount").val(ui.value);
    }
});

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I guess your name and is are different for the textbox..Make it same and if you have more than one textbox change the name of the textbox into array otherwise you will get only one value after posting it.

from which name do i post it?

That i have added in the answer.I have changed the name and id of your textbox into jquery_text

I don't think it is with the naming, it was on the order in which things took place. The submission happened before the content append. Check my solution, it worked for me =)

It is good idea.. I tried in your way. but, i'm getting only static element value in php. I could not get dynamic element values.

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php - how do i post textbox value which is dynamically created by jque...

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Now it's working for 50%. If i change on slider it changes on select, but it doesn't work when i change values on select.

<script> 
    $(function() {
        var select = $( "#czastrwania" );
        var slider = $( "<div id='slider'></div>" ).insertAfter( select ).slider({
        min: 60,
        max: 240,
        step: 30,
        range: "min",
value: select[ 0 ].selectedIndex + 1,
            slide: function( e, ui ) {
            var hours = Math.floor(ui.value / 60);
            var minutes = ui.value - (hours * 60);

            if(minutes == 0) minutes = '00';
            select[ 0 ].value = hours+':'+minutes;
            }
        });
        $( "#czastrwania" ).change(function() {
            slider.slider( "value", select[ 0 ].value );
        });
    });
    </script>

php - Problem with jQuery slider - time - Stack Overflow

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Make sure $('.middle_container').find('.start_price').val() and $('.middle_container').find('.end_price').val() are returning proper values. Also to set the value of the slider you have to use the same syntax which you are using for setting min/max values. Also

$.get("ajax.php",options,function(data){
    $('.middle_container').html(data);          

    $('#slider-range').slider( "option", "min", $('.middle_container').find('.start_price').val() );
    $('#slider-range').slider( "option", "max", $('.middle_container').find('.end_price').val() );
    $('#slider-range').slider( "option", "value", $('#slider-range').slider("value"));

});

how do i access each of the values separately? because this is a range so there are two handles

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PHP is used to render the page, once the page is rendered php can no longer interact with it. Using jquery you can make an ajax post back to your site with the 2 values something like:

var slider1FinalValue = false;
var slider2FinalValue = false;

$( "#slider1" ).slider({
  stop: function( event, ui ) {
    slider1FinalValue = $(this).slider( "value" );
    sendBothValuesIfSet();
  }
});
$( "#slider2" ).slider({
  stop: function( event, ui ) {
    slider1FinalValue = $(this).slider( "value" );
    sendBothValuesIfSet();
  }
});
function sendBothValuesIfSet(){
  if(slider1FinalValue !==false && slider2FinalValue !==false){
    $.post(
        'www.mydomain.com/myscript.php',
        {slider1:slider1FinalValue, slider2:slider2FinalValue }
    );
  }
}

Then in your php file:

<?php
if($_POST && !empty($_POST)){
  $sliders = json_decode($_POST);
  $multipliedValue = $sliders['slider1'] * $sliders['slider2'];
}

Something along those lines. I havent tested this code just wrote it on the fly so if you have issues let me know and ill adjust it as needed.

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$( "#slider-range2" ).slider({
    range: true,
    min: 18,
    max: 90,
    values: [ <?php echo $value1; ?>, <?php echo $value2; ?> ],
    slide: function( event, ui ) {
        $( "#amount2" ).val( ui.values[ 0 ] + "-" + ui.values[ 1 ] );
  }

where of course $value1 and $value2 are the values you get from the database.

-edit- or what the other guy said, you can use a php for loop to make the size of your array dynamic that way :)

is a .js file, i cant write php code there.

@alejoabella Of course you can, just change the file extension.

maybe then you could also consider creating a php web method and call that from your jquery code?

@mblase75 completely forgot that the browser is not concerned about the extension. However doesn't it not cache the file like a normal js but the php is dynamic?

@mblase75 thanks, i know but is not the idea.

How to pass PHP variable value to jquery function? - Stack Overflow

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$( "#slider-range2" ).slider({
    range: true,
    min: 18,
    max: 90,
    values: <?php echo $your_variable ?> ,
    slide: function( event, ui ) {
        $( "#amount2" ).val( ui.values[ 0 ] + "-" + ui.values[ 1 ] );
    }
});

is a .js file, i cant write php code there.

How to pass PHP variable value to jquery function? - Stack Overflow

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When you dynamically create the textboxes give them both an unique id tag and a name tag (same values)

You'll then be able to access them from JS with

$('#TB' + n).val()

where n is the text box number you want to access.

As they are in a form, the values will be posted to the speciefied form url in the tag action and will be available for processing with PHP

OP is already doing what you suggest, however it lacks method to prevent duplicating ID's across forms. Selector you are using is invalid (missing prefix)

Hi @charlietfl, in your code is the array you are pulling data from using PHP called updateMasterNumbersArray()? Your code disabled the "Refresh all sets" button function for some reason. Also, your check function of duplicate values is occurring with all the textbox fields in every number set and checking for duplication. What I was seeking was to check each number set of 3 numbers individually and not checking against the other number sets. Each set of 3 numbers are checked for duplicates but not against the other number sets. Is that possible to do? Thanks much.

ahhh.. wish I'd known about only check row for duplicates. That is so much easier and will cut out a bunch of my code. As for php , the javascript you see has nothing to do with php. The only php I've used is print_r($_GET) in a file that is the action for form. I'll revise the duplicate checks to just rows and fix the validation to match. Also will fix the reset

Hi, @charlietfl, were you able to make the changes so each set (or row) of textboxes can be checked for duplicate numbers if the user changes them and making the reset function again? Do you know how to add a button dynamically for each set of textboxes (row) to clear the textboxes and allow for new numbers to be generated in them? Also, can you provide your PHP code for how you submit the textbox fields for inserting into MySQL? My PHP code does not seem to be working. Thanks much.

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Ok I changed the way this works and I was trying to display as tab section on reload I think before it had been dynamically created. I've now removed that and jquery appears to be loading ok now. So the user will have to select an option from the dropdown but for now that is ok.

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Destroy the slider first, which removes the slider functionality completely. This will return the element back to its pre-init state.

$("#selector").slider("destroy");

After that you can add new values to the slider as,

$("#selector").slider({
    range: "max",
    min: 0, // min value
    max: 200, // max value
    step: 0.1,
    value: 200, // default value of slider
    slide: function(event, ui) {
        $("#amount").val(ui.value);
    }
});

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Not 100% sure what you are asking, but if you need to set the range values programmatically based on a server response, then when you get the data from the server, you can get/set the values using the jQuery UI getter/setter.

The following shows the example of getting/setting the range, values, and value

//getter
var range = $( ".selector" ).slider( "option", "range" );
//setter
$( ".selector" ).slider( "option", "range", 'min' );

//getter
var values = $( ".selector" ).slider( "option", "values" );
//setter
$( ".selector" ).slider( "option", "values", [1,5,9] );

//getter
var value = $( ".selector" ).slider( "option", "value" );
//setter
$( ".selector" ).slider( "option", "value", 37 );

php - Setting Range for jQuery UI Sliders Based on Results of SQL Quer...

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You could use ajax to pass the slider values to the PHP script.

Say you store the slider values as below:

$value = $("#slider").slider("value"); //or however you get the value

Then you can perform a post call like this:

$.post("test.php", { $slider_value: $value } );

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If you want to do this on an input value changing you can do something like this.

$('#inputid').focusout(function() {

    // slider destroy and create as Shailesh showed
});

I was going to use keyup instead of focus, but depending on the internal checks you make, you could end up rebuilding your slider many times which would not be a benefit as you wont use it until you moving on from the input.

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you are creating your variable in javascript and call through php its not possible, you need to use

document.getElementById("myDiv").innertHTML = minValue+maxValue
<div id="leftSlider"></div> <!-- display result here -->

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your code is not very clear: you seem to have put everything inside a php print call? That's not a very good idea. Instead only open the <\?php tag when you really need dynamic content from the server. To answer your specific question, all you need to do is echo or print your php variable, just like you would when dynamically creating an html id for instance from a php variable:

<script type="text/javascript">

function vediticket(ID){
  $.ajax({
            type: "POST",
            url: "inc.miapagina.inc.php",
            data: "idric=" + <?php echo $idrichiesta;?>,

hum correction sorry... obviously the php echo should be in the function call, not the function definition of the inconvenience of doing two things at once: so keep your original code and when you call the function do:

vediticket(<?php echo $idrichiesta;?>);

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$slider->setJQueryParams(array(
  'min' => 0,
  'max' => 60,
  'value' => 15,
  'slide' => new Zend_Json_Expr('function(event, ui) {
    $("#amount").val('$' + ui.value);
   }')
));

php - Zend_Form jQuery slider - Stack Overflow

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Change the jQuery code to this, and it should work fine. Your problem was that you were adding the contents with the "submit" button's click, which would submit the form and the contents would get appended after the submit request.

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if it's running ok in other browsers you can use timers and manuplate the min and max values accordingly .

$( "#price-range" ).slider({
range: true,
min: min,
max: max,
values: [ value1, value2 ],
slide: function( event, ui ) {

    $('#min').val(ui.values[ 0 ]);
    $('#max').val(ui.values[ 1 ]);

    $('#order').val($("#price").val());
      // you may use the timer over here and submit the form only 
   //once not every time.. Last submution would be ok but dont forget to clear timer on previous slidings..
    $('#filter-form').submit();

}
 });

php - jquery slider issue with mozilla - Stack Overflow

php google-chrome jquery-ui slider mozilla