How to access userfile names?

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  • Don
    #1

    How to access userfile names?



    The following 4 statements don't work. What is the proper way to
    pickup the filenames corresponding to each of the userfiles in the
    "input" tags? Thanks for your help. --Don

    Following is part of a page that runs on the client side:
    -----------------------------------------------------------------
    var picture1 = document.forms. postamessagedat a.userfile.name[0];
    var picture2 = document.forms. postamessagedat a.userfile.name[1];
    var picture3 = document.forms. postamessagedat a.userfile.name[2];
    var picture4 = document.forms. postamessagedat a.userfile.name[3];

    Following is part of a form to be submitted to PHP upload processor
    below. This resides in the same page as the foregoing.
    --------------------------------------------------------------------------------
    <p>Optional picture #1:<br><input type=file name="userfile[]" size=90
    maxlength=20000 0 tabindex=3>;
    <p>Optional picture #2:<br><input type=file name="userfile[]" size=90
    maxlength=20000 0 tabindex=3>;
    <p>Optional picture #3:<br><input type=file name="userfile[]" size=90
    maxlength=20000 0 tabindex=3>;
    <p>Optional picture #4:<br><input type=file name="userfile[]" size=90
    maxlength=20000 0 tabindex=3>;

    Following processes file uploads and resides on server:
    -----------------------------------------------------------------
    <?php
    $uploaddir = 'picturevault/';
    $uploadfile1 = $uploaddir . $_FILES['userfile']['name'][0];
    $uploadfile2 = $uploaddir . $_FILES['userfile']['name'][1];
    $uploadfile3 = $uploaddir . $_FILES['userfile']['name'][2];
    $uploadfile4 = $uploaddir . $_FILES['userfile']['name'][3];

    print "<pre>";

    if ($_FILES['userfile']['size'][0] != 0)
    move_uploaded_f ile($_FILES['userfile']['tmp_name'][0],
    $uploadfile1);

    if ($_FILES['userfile']['size'][1] != 0)
    move_uploaded_f ile($_FILES['userfile']['tmp_name'][1],
    $uploadfile2);

    if ($_FILES['userfile']['size'][2] != 0)
    move_uploaded_f ile($_FILES['userfile']['tmp_name'][2],
    $uploadfile3);

    if ($_FILES['userfile']['size'][3] != 0)
    move_uploaded_f ile($_FILES['userfile']['tmp_name'][3],
    $uploadfile4);

    print "</pre>";

    ?>




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    • lallous
      #2
      Re: How to access userfile names?

      Hello

      You will find this useful:

      <html>
      <head>
      <title> New Document </title>
      <meta name="Generator " content="EditPl us">
      <meta name="Author" content="lallou s">
      <meta name="Keywords" content="">
      <meta name="Descripti on" content="">
      </head>

      <script language="JavaS cript">
      <!--
      function f1()
      {
      var f = document.f, e, s = '';
      for (i=0;i<f.elemen ts.length;i++)
      {
      e = f.elements[i];
      x = e.name;
      e = f.elements[i];
      s += "i=" + i + " elements name='" + x + "' element value='" + e.value +
      "'\n";
      }
      alert(s);
      }
      //-->
      </script>
      <body onload="f1();">
      <form name="f">
      <input type="text" name="f_t[]" value="1">
      <input type="text" name="f_t[]" value="2">
      <input type="text" name="f_t[]" value="3">
      <input type="text" name="f_t[]" value="4">
      </body>
      </html>


      p.s: you may replace the type="text" w/ type="file".
      And to access its value you do it as: elementName.val ue.

      --
      Elias

      "Don" <no@adr.com> wrote in message
      news:lkdfl0pu91 ugs64e7dkbbp70r bm6ckvnr3@4ax.c om...[color=blue]
      >
      >
      > The following 4 statements don't work. What is the proper way to
      > pickup the filenames corresponding to each of the userfiles in the
      > "input" tags? Thanks for your help. --Don
      >
      > Following is part of a page that runs on the client side:
      > -----------------------------------------------------------------
      > var picture1 = document.forms. postamessagedat a.userfile.name[0];
      > var picture2 = document.forms. postamessagedat a.userfile.name[1];
      > var picture3 = document.forms. postamessagedat a.userfile.name[2];
      > var picture4 = document.forms. postamessagedat a.userfile.name[3];
      >
      > Following is part of a form to be submitted to PHP upload processor
      > below. This resides in the same page as the foregoing.
      > --------------------------------------------------------------------------------
      > <p>Optional picture #1:<br><input type=file name="userfile[]" size=90
      > maxlength=20000 0 tabindex=3>;
      > <p>Optional picture #2:<br><input type=file name="userfile[]" size=90
      > maxlength=20000 0 tabindex=3>;
      > <p>Optional picture #3:<br><input type=file name="userfile[]" size=90
      > maxlength=20000 0 tabindex=3>;
      > <p>Optional picture #4:<br><input type=file name="userfile[]" size=90
      > maxlength=20000 0 tabindex=3>;
      >
      > Following processes file uploads and resides on server:
      > -----------------------------------------------------------------
      > <?php
      > $uploaddir = 'picturevault/';
      > $uploadfile1 = $uploaddir . $_FILES['userfile']['name'][0];
      > $uploadfile2 = $uploaddir . $_FILES['userfile']['name'][1];
      > $uploadfile3 = $uploaddir . $_FILES['userfile']['name'][2];
      > $uploadfile4 = $uploaddir . $_FILES['userfile']['name'][3];
      >
      > print "<pre>";
      >
      > if ($_FILES['userfile']['size'][0] != 0)
      > move_uploaded_f ile($_FILES['userfile']['tmp_name'][0],
      > $uploadfile1);
      >
      > if ($_FILES['userfile']['size'][1] != 0)
      > move_uploaded_f ile($_FILES['userfile']['tmp_name'][1],
      > $uploadfile2);
      >
      > if ($_FILES['userfile']['size'][2] != 0)
      > move_uploaded_f ile($_FILES['userfile']['tmp_name'][2],
      > $uploadfile3);
      >
      > if ($_FILES['userfile']['size'][3] != 0)
      > move_uploaded_f ile($_FILES['userfile']['tmp_name'][3],
      > $uploadfile4);
      >
      > print "</pre>";
      >
      > ?>[/color]


        Comment

        • Grant Wagner
          #3
          Re: How to access userfile names?

          lallous wrote:
          [color=blue]
          > Hello
          >
          > You will find this useful:
          >
          > <html>
          > <head>
          > <title> New Document </title>
          > <meta name="Generator " content="EditPl us">
          > <meta name="Author" content="lallou s">
          > <meta name="Keywords" content="">
          > <meta name="Descripti on" content="">
          > </head>
          >
          > <script language="JavaS cript">[/color]

          <script type="text/javascript">

          The language attribute is deprecated.
          [color=blue]
          > <!--[/color]

          Not required or desired.
          [color=blue]
          > e = f.elements[i];
          > x = e.name;
          > e = f.elements[i];[/color]

          I'm not sure why you're grabbing a reference to the form element twice.
          [color=blue]
          > //-->[/color]

          Not required or desired.
          [color=blue]
          > </script>
          > <body onload="f1();">
          > <form name="f">[/color]

          You are missing a closing </form> tag.
          [color=blue]
          > </body>
          > </html>[/color]

          <form>
          <p>Optional picture #1:<br><input type=file name="userfile[]" size=90
          maxlength=20000 0 tabindex=3>;
          <p>Optional picture #2:<br><input type=file name="userfile[]" size=90
          maxlength=20000 0 tabindex=3>;
          <p>Optional picture #3:<br><input type=file name="userfile[]" size=90
          maxlength=20000 0 tabindex=3>;
          <p>Optional picture #4:<br><input type=file name="userfile[]" size=90
          maxlength=20000 0 tabindex=3>;

          Read the values with:
          <input type="button"
          value="Show picture names (example only)"
          onclick="showIn putValuesOf(thi s.form, 'userfile[]');">
          </form>

          <script type="text/javascript">
          function showInputValues Of(f, inputName) {
          var output = [];
          var inputs = f.elements[inputName];
          if (typeof inputs.length == 'undefined') {
          inputs = [ inputs ];
          }
          for (var i = 0; i < inputs.length; i++) {
          output.push(inp uts[i].value);
          }
          alert(output.jo in('\n'));
          }
          </script>

          --
          Grant Wagner <gwagner@agrico reunited.com>
          comp.lang.javas cript FAQ - http://jibbering.com/faq

            Comment

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