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jumbo125

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  1. Hy, I tried to read filenames of zip file. I use the jszip libaray to see the fileneams and check name and type, BUT the libaray can't show me the size of files. or i don't know how i can see it. Does anybody knows a zip libary, which is easy to use? i don't want compress and uncompress, only read the files and their properties..... thank you
  2. No, normale i would Write Print_r($_FILES) and Show what the php file gives out. That is the reason, for my last question. I didn't know how i can see the Output when the Site doesn't change
  3. Thank you for answer. I know how i get upload files via php like:$_FILES [inpuzname]But i Never Try to get files which are uploaded with jsThe other Problem is, that i Write with this file some Database entrys....It will be really nice to See some errorsI will Try the error massage postsSorry for my wr
  4. // Edit: i try it with an multiple array, formData("name", "array"); array( [0] => array [0] => inputname e.g. field_1 [1] => array mit filename [0] => filnema 1 [1] => filname2 [1] => array [0] => inputname e.g. field_2 [1] => array with filenames from second input type [0] => filename1 [1] => filename2 ) client.send(formData); If this work, i have only ONE Problem. i don_t can see the result, because it doesn't change to the php file.... omg, what can I do.... I click on "send" and get only one sentence in the console, but i don't know if there is an fail in the php file, or something more. what can i do to change the site??? i want to get out the POST files eand more. I think it's reallay difficult to understand my Problem. with the normall send method, without js submit buttons and action location i changed the site. than i saw the files with echo and print_ and can create the perfect php file. Now i don't know, what i should wirte in the php file.
  5. I know the method appendI don't understand how i can send the formDataFor exampleFormData1.append("file1","filelist1")FormData2.append ("file2", "filelist2")And more... Maybe 10 and more...But how i can send alltogether Client.send(formdata1,formdata3)
  6. Thank you very michNow i know how i can put the multiple files in the var file.But i don't know how i use the other input tags. I put the var file in the formdata and send it with the Client.How can i send the other input tags? Can i create more formdata and send all formdata with one client
  7. I try to create an Upload with javascript nad progress bar. i found this code in the internet, but i can't change the code for my idea. This is the script sample: var client = null; function uploadFile(){ //Wieder unser File Objekt var file = document.getElementById("fileA").files[0]; var formData = new FormData(); client = new XMLHttpRequest(); if(!file) return; formData.append("datei", file); client.onerror = function(e) { alert("onError"); }; client.open("POST", "upload.php"); client.send(formData);} my first problem is, that the autor used ONE input tag an set no MULTIPLE. I have a multiple input tag. I think it's enoug to change: var file = document.getElementById("fileA").files[0]; to var file1 = document.getElementById("file1").files; //dlete [0] for all files and formData.append("datei", file); to formData.append("file1[]", file);// add [] to create an array is this correct? AND MY IMPORTANT PROBLEM is, that i have more input tags: <form> <input type="file" name="file1"> <input type="file" name="file2"> <input type="file" name="file3"> </form> How can i pass the code for mulitple tags with different names??????? I search since 4 days for an idea. It will be perfect, if i can send all input with ONE clinet. IT's really important to read the files with $_FILES[file1], $_FILES[file2]
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