1. In this tutorial we are going to tell you that how to send an image through ajax to process.<\/p>\n\n\n\n
2. In the following code you can process image as soon as it is selected.<\/p>\n\n\n\n
3. First You need to Select image file through following line of html code<\/p>\n\n\n\n
<input type=\"file\" id=\"image_to_upload\"\/><\/pre>\n\n\n\n4. After that you need to write the following jQuery code to catch this event.<\/p>\n\n\n\n
<script type=\"text\/javascript\">\n\tjQuery.noConflict();\t\n\tformdata = new FormData();\t\t\n\tjQuery(\"#image_to_upload\").on(\"change\", function() {\n\t\tvar file = this.files[0];\n\t\tif (formdata) {\n\t\t\tformdata.append(\"image\", file);\n\t\t\tjQuery.ajax({\n\t\t\t\turl: \"destination_ajax_file.php\",\n\t\t\t\ttype: \"POST\",\n\t\t\t\tdata: formdata,\n\t\t\t\tprocessData: false,\n\t\t\t\tcontentType: false,\n\t\t\t\tsuccess:function(){}\n\t\t\t});\n\t\t}\t\t\t\t\t\t\n\t});\t\n<\/script><\/pre>\n\n\n\n5. In the destination ajax file you can access image by the following variable<\/p>\n\n\n\n
$_FILES[\"image\"][\"name\"]<\/pre>\n\n\n\n6.That is it Save And Enjoy!!<\/p>\n","protected":false},"excerpt":{"rendered":"