Changing Image Based On Selection In 2 Dropdowns

I have two select elements: ab<Solution 1: You need to write a function which is called on the change event of both select elements which concatenates the image filename and updates the src property of the required img, something like this:$('select.form-control').change(function() { var filename = $('#shape').val() + '-' + $('#waist').val() + '.jpg'; $('#imgToChange').prop('src', filename); }); CopySolution 2: You can use this to get the combined value of dropdowns in the format like a-2,a-3,b-1 etc.document.getElementById("shape").value + "-" + document.getElementById("waist").valueCopySolution 3: you can use the below plunker as well.https://plnkr.co/edit/1P12J0Ahl2S1m6aJ1rD0?p=preview<selectclass="form-control"id="shape"name="shape"><optionvalue=""></option><optionvalue="a">a</option><optionvalue="b">b</option><optionvalue="c">c</option><optionvalue="d">d</option><optionvalue="e">e</option><optionvalue="f">f</option></select><selectclass="form-control"id="waist"name="waist"><optionvalue=""></option><optionvalue="1">1</option><optionvalue="2">2</option><optionvalue="3">3</option></select><buttononClick="test()">Proceed</button>Copyin script.var test = function(){ if($("#shape").val() != "" && $("#waist").val() != "") { alert("Update"); } else { alert("select both the option"); } }; Copyhere each select box has the default blank option and if user try to proceed then the value is checked and appropriate message is displayed.Solution 4: You can use a solution with jQuery (javascript) / PHP / AJAX (and don't worry, AJAX is easy). It will work like this:jQuery - detect when both SELECT controls have been changed, AJAX - a jQuery block that sends data to a back-end PHP file, PHP - Receives AJAXed data and identifies the correct image (e.g. from MySQL) then builds HTML code and ECHOs it back to jQuery side, AJAX - Receives PHP data via AJAX, jQuery - Inside the AJAX success function, uses the ,html() method to insert the new HTML into the DOMThat's the overview. Here's how it will look code-wise:HTML:<selectclass="form-control"id="shape"name="shape"><optionvalue="">Options go here</option></select><selectclass="form-control"id="waist"name="waist"><optionvalue="">Options go here</option></select><divid="someDIV"> //DIV that will receive IMG tag </div>CopyjQuery:var chg = false; $('#shape, #waist').change(function(){ if (!chg){ chg=true; }else{ var myimg = $('#shape').val() + $('#waist').val(); $.ajax({ type: 'post', url: 'my_second_php_file.php', data: 'myimg=' +myimg, success: function(d){ //if (d.length) alert(d); $('#someDIV').html(d); } }); //END AJAX }//end else }); //END select.changeCopyPHP:<?php$img = $_POST['myimg']; //echo ($img); die();//$result = Do your MySQL lookup here$out = ' <img src="' .$result['img']. '" class="yourImage" /> '; echo$out; ?>CopyHere is a jsFiddle DEMO that approximates how it will work. jsFiddle cannot do AJAX, so the AJAX routine is replaced by something that will demonstrate how it will look, rather than a true working example.Here are some simple demos of AJAX code and a brief explanation -- copy them to your own system and experiment. Really quite simple, and xtreme useful.Here are some free, 10-min video mini-tuts on AJAX. Share Post a Comment for "Changing Image Based On Selection In 2 Dropdowns" Top Question How To Validate Form Fields And Post The Form Data To Server Using Jquery And Ajax? 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