php - Inserting Value into database of hidden fields -
query if select 'select new parent category' dropdown, new textbox gets appear. wants insert value database. textbox hidden bydefault.
html file
<head> <script type="text/javascript"> $(document).ready(function() { $('#parentcat').change(function(){ if($('#parentcat').val() === 'create new parent category') { $('#other').show(); $('#catname').hide(); } else { $('#other').hide(); $('#catname').show(); } }); }); </script> </head> <table> <body> <form method="post" action="" name="form1" enctype="multipart/form-data" > <tr> <td><h4>add category</h4></td> </tr> <tr> <td>select parent category:</td><td><select name="parentcat" id="parentcat"> <?php include "../storescripts/connect_to_mysql.php"; $result=mysql_query("select parent category"); while($row=mysql_fetch_array($result)){ echo "<option>".$row['parent'];"</option>"; } ?> <option value="create new parent category">create new parent category</option> </select><br/></td> <tr><td><input type="text" id="other" name="other" placeholder="new parent category name:" style="display: none;"/></td></tr> </tr> <tr> <td></td><td><input type="text" name="catname" placeholder="subcategory name" id="catname"><br/></td> </tr> <tr> <td>status: </td> <td><select name="status" > <option selected="active">active</option> <option>inactive</option> </select></td> </tr> <tr><td>category image :</td><td><input type="file" name="imageupload" id="imageupload"></td> </tr> <tr> <td></td> <td><input type="submit" name="submit" value="insert"></td> </tr> </form> </table> here complete html file,bydefault 'other' dropdown hidden. here i'm providing php code inserts values database, issue how can insert value of hidden filed database. database fields includes : cat_id(auto-increment),cat_name,parent,cat_status,cat_image. here php code inserting values database.
<?php if(isset($_post['submit'])) { $parentcat=$_post['parentcat']; $catname=$_post['catname']; $status=$_post['status']; $target_dir = "../uploads/"; $target_file = $target_dir . basename($_files["imageupload"]["name"]); $uploadok = 1; $imagefiletype = pathinfo($target_file,pathinfo_extension); if (move_uploaded_file($_files["imageupload"]["tmp_name"], $target_file)) { echo "the file ". basename( $_files["imageupload"]["name"]). " has been uploaded."; } else { echo "sorry, there error uploading file."; } $image=basename( $_files["imageupload"]["name"],".jpg"); // used store filename in variable $sql = mysql_query("select cat_id category cat_name='$catname' limit 1"); $catmatch = mysql_num_rows($sql); // count output amount if ($catmatch > 0) { echo 'sorry tried place duplicate "category name" system, <a href="addcategoryy.php">click here</a>'; exit(); } else{ $sql=mysql_query("insert `category` (`cat_name`, `parent`, `cat_status`,`cat_image`) values ('$catname', '$parentcat', '$status','$image') "); echo $sql; } } ?> for eg. if select new parent category subcategory named textbox appears , want insert value database. how can that.
change jquery code this
<script type="text/javascript"> $(document).ready(function () { $('#parentcat').change(function () { if ($('#parentcat').val() === 'create new parent category'){ $('#other').prop('type', 'text'); $('#catname').prop('type', 'hidden'); } else{ $('#other').prop('type', 'hidden'); $('#catname').prop('type', 'text'); } }); }); </script> also change html code #other input box this
<input type="hidden" id="other" name="other" placeholder="new parent category name:"/> let me know if works you.
edit
in php file, instead of $catname=$_post['catname']; add this
$catname = ( $parentcat == "create new parent category") ? $_post['other'] : $_post['catname'];
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