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PHP - Dynamic Search Form For Mysql Database
I'm using some code to create a select menu of a fieldname of data I have in a mysql database :
Code: [Select] <?php mysql_connect('localhost' , 'dbname', 'password'); mysql_select_db('dbname'); $result=mysql_query("SELECT * FROM Persons"); if(mysql_num_rows($result)>0) { ?> <select name="Persons"> <?php while($rows=mysql_fetch_array($result)){ ?> <option value="<?php echo $rows['id']; ?>"> <?php echo $rows['FirstName']; ?></option> <?php } ?> </select> What I would like to do is to elevate this into a jump menu form so that if the user selects an item from my form they are taken to a results page showing the full row of data from the database. Basically a search form containing items from the database they can choose to see more details on. eg. In my example you select a persons name and then you are taken to a results page which displays the details of that person from the database. Problem is I don't know how to do this and have been trawling around for a couple of days to find a solution (sorry I'm new to php). I would appreciate some help or a working example would be great of : 1/ A working dynamic jump menu 2/ The page that would process the form 3/ The results page displaying the data I have selected. Thank you for your time... Similar TutorialsThis topic has been moved to Ajax Help. http://www.phpfreaks.com/forums/index.php?topic=326703.0 Need some help I have 2 tables in a database and I need to search the first table and use the results from that search, to search another table, can this be done? and if it can how would you recommend that I go about it? Thanks For Your Help Guys! im making a game and i need to show a users money but i dont know how help? I can not get the values from the javascript add row to go dynamically as a row into MySql only the form values show up as the form below as one row. I made it as an array, but no such luck, I have tried this code around a multitude of ways. I don't know what I am doing wrong, kindly write out the correct way.
<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR...ransitional.dtd"> <html xmlns="http://www.w3.org/1999/xhtml"> <head> <meta http-equiv="Content-Type" content="text/html; charset=utf-8" /> <title>Dynamic Fields js/php to MySql need to submit dynamically to the database</title> <?php require ('database.php'); ?> <script type="text/javascript"> var counter = 1; var collector = ""; function addfields(indx) { var tbl = document.getElementById('table_id'); var newtr = document.createElement('tr'); counter = counter + indx; newtr.setAttribute('id','tr'+counter); newtr.innerHTML = '<td><input type="checkbox" name="checkb'+counter+'" id="checkb'+counter+'" value="'+counter+'" onclick="checkme('+counter+')"></td><td><input type="text" name="text1[]"></td><td><textarea name="textarea1[]"></textarea></td>'; tbl.appendChild(newtr); } function checkme(dx) { collector += dx+","; } function deletetherow(indx) { var col = collector.split(","); for (var i = 0; i < col.length; i++) { var remvelem = document.getElementById('tr'+col[i]); var chckbx = document.getElementById("checkb"+col[i]); if(remvelem && chckbx.checked) { var tbl = document.getElementById('table_id'); tbl.removeChild(remvelem); } } } </script> </head> <body> <form enctype="multipart/form-data" id="1" style="background-color:#ffffff;" action="<?php echo $_SERVER['PHP_SELF']; ?>"></form> <table id="table_id" > <tr id="tr1" class="trmain"> <td> </td> <td> <input type="text" name="text1[]"> </td> <td> <textarea name="textarea1[]"></textarea> </td> </tr> </table> <input type="button" value="Add" onClick="addfields(1);" /> <input type="button" value="Delete" onClick="deletetherow()" /> <input type="submit" value="Send" id="submit" name="submit"/> <?php if(isset($_POST['submit'])) { for ($i=0; $i < count($_POST['text1']); $i++ ) { $ced = stripslashes($_POST['text1'][$i]); $erg = stripslashes($_POST['textarea1'][$i]); } $bnt = mysql_query("INSERT INTO tablename (first, second) VALUES ('$ced', '$erg')")or die('Error: '. mysql_error() ); $result = mysql_query($bnt); } ?> </body> </html> Hello, so i made table with Tags row which is Text(126).. and idea is to put text like "PHP, MySQL, Forum, Blog" etc etc...
I have problem searching tags my search code:
$requestedtag = $con->real_escape_string($_GET['tag']);
$sql = "SELECT * FROM images WHERE tags = '$requestedtag' ORDER BY ID DESC LIMIT 12";
$result = $con->query($sql);
if ($result->num_rows > 0)
{
while($row = $result->fetch_assoc()) {
$image_id = $row["ID"];
$image_title = $row["Title"];
$image_url = $row["ImgURL"];
$image_tags = $row["Tags"];
echo '<div class="col-3"><div class="thumb"><a href="/image.php?id='.$image_id.'" title="'.$image_title.'"><img src="'.$image_url.'"></img></a></div></div>';
}
}
But it doesnt return anything. I think the problem is that row["Tags"]; is divided by "," from tags and i should split it somehow but i dont really know what to do...
I have a search set up to search a table for the text entered in a textbox, I have two columns in the table, one with the first name of people, and the second with their last names, I am wondering how I can search both, so for instance: I type in the search field: Roger Smith in the database it would look like: First_name-----|-----Last_name -------------------|------------------- Roger------------|-------Smith my current query is: Code: [Select] $query = mysql_query("SELECT * FROM users WHERE fname LIKE '%$find%' OR lname LIKE '%$find%'"); But if I type both parts of the name it doesn't return anything. works fine if I just search for "Roger" OR "Smith". Hi, I'm am trying to use a search engine that search's for a username in the database and displays the information back. I have searched for a script but none of them has helped me. I have tried to use this without any luck: mysql_connect("localhost", "", "") or die(mysql_error()); mysql_select_db("dbsystem") or die(mysql_error()); $todo=$_POST['todo']; if(isset($todo) and $todo=="search"){ $search_text=$_POST['search_text']; $type=$_POST['type']; $search_text=ltrim($search_text); $search_text=rtrim($search_text); if($type<>"any"){ $query="select * from users where name = '$search_text'"; }else{ $kt=split(" ",$search_text);//Breaking the string to array of words // Now let us generate the sql while(list($key,$val)=each($kt)){ if($val<>" " and strlen($val) > 0){$q .= " name like '%$val%' or ";} }// end of while $q=substr($q,0,(strlen($q)-3)); // this will remove the last or from the string. $query="select * from users where $q "; } // end of if else based on type value echo $query; echo "<br><br>"; $nt=mysql_query($query); echo mysql_error(); while($row=mysql_fetch_array($nt)){ echo "$row[name]<br>"; } // End if form submitted }else{ echo "<form method='post' action=''><input type='hidden' name='todo' value='search' /> <input type='text' name='search_text' /><input type='submit' value='Search' /><br> <input type='radio' name='type' value='any' checked />Match any where <input type='radio' name='type' value='exact' />Exact Match </form> "; } I will be grateful if you can help with this. Hi. I have a database named 'Forums' with a table named 'forumlist'. The fields in the table are 'id', 'name', 'link', and 'keys'. I am trying to search the database in either the 'name' or 'keys' field and display the matches. I am getting this error: Parse error: syntax error, unexpected T_STRING, expecting ']' in C:\xampp\htdocs\search3.php on line 51 (The define fields) <html> <head> <title>My Forum Search</title> <style type="text/css"> table { background-color: #CCC } th { width: 150px; text-align: left; } </style> </head> <body> <h1>Forum Search</h1> <form method="post" action="search3.php"> <input type="hidden" name="submitted" value="true" /> <label>Search Category: <select name="category"> <option value="keys">Keyword</option> <option value="name">Name</option> </select> </label> <label><input type="text" name="criteria" /></label> <input type="submit" /> </form> <?php if (isset($_POST['submitted])) { DEFINE ('DB_USER', 'root'); DEFINE ('DB_PSWD', '******'); DEFINE ('DB_HOST', 'localhost'); DEFINE ('DB_NAME', 'Forums'); $dbcon = mysqli_connect(DB_HOST, DB_USER, DB_PSWD, DB_NAME); $category = $_POST['category']; $criteria = $_POST['criteria']; $query = "SELECT * FROM forumlist WHERE $category LIKE '$criteria'"; $result = mysqli_query($dbcon, $query) or die ('Error retrieving data') echo "<table>"; echo "<tr> <th>Name</th><th>Link</th> </tr>"; while ($row = mysqli_fetch_array($result, MYSQLI_ASSOC)) { echo "<tr><td>"; echo $row['name']; echo "</td><td>"; echo $row['link']; echo "</td></tr>"; } echo "</table>"; } // end of main if state ?> </body> </html> Thanks in advance I'm praying there's a function somewhere out there that: generates a html form based on the fields of a particular table within a DB (namely mySQL) generates the appropriate form input based on the fields type or notes (ie. for field: avatar_img, it knows to generate a file input) uses the POST method to submit generates and submits SQL insert query to database In addition to that, a similar function that generates a form that allows the editing of records. It doesn't take too long to write these each time myself, but if there's a function that you provide the formname, database driver, table name, whether your form is to insert or edit, that would be quite nifty i think. ie. generateForm(addUser, mySQL, tUsers, insert) Seen any such thing? Many thanks. Say a user puts in a support request, and for every request it generates a unqiue string, and enters it into the database. Ok, now say there is a text field, when the user enters their unique string and it finds a match, it displays the data along with it. How can I accomplish this? Im kind of new to mysql, but I know basic SQL. Would be great if somebody could point me in the right direction! Thanks Hey guys, Hi, Hoping someone could help me, im not great at php programming and im trying to implement a search form which searches a sql database but for some reason its not working i keep on getting the following error.. " Warning: mysql_fetch_array() expects parameter 1 to be resource, boolean given in C:\xampp\htdocs\site\contact\search.php on line 141" Im not sure why this happens and its really starting to bug me now :/ I have to pages to the search feature, one page which contains the <form> and the other with the PHP code.... the form.. <form method="get" action="search.php"> <label>Search Term: <input type="text" id="query" name="query"/> </label> <label> <select name="category" id="category"> <option value="none">Please Select a Type</option> <option value="teamname">Manager Name</option> <option value="manager">Team Name</option> <option value="age">Age Group</option> </select> </label> <input name="Search" type="submit" value="Search"/> </form> the PHP code.. <?php $query=$_GET['query']; $db_host="localhost"; $db_username="root"; $db_password=""; $db_name="info"; $db_tb_name="info"; $db_tb_atr_name=$_GET['category']; mysql_connect("$db_host","$db_username","$db_password"); mysql_select_db("$db_name"); $query_for_result=mysql_query("SELECT * FROM $tb_name WHERE $db_tb_atr_name like '%".$query."%'"); while($data_fetch=mysql_fetch_array($query_for_result)) { echo "<p>"; echo $data_fetch['table_attribute']; echo "</p>"; } mysql_close(); ?> Id be really great full for any light you guys could shed on this for me with thanks, fozze In brief, I'm attempting to capture the form data from a dynamic form to a mysql database. From he
To He
The post data looks like this from the form - Array(
... ) PHP CODE:
... As a non-php or mysql developer, I'm learning on the fly. This is a section of the php file that I was using to post the form data to mySQL. This worked great up to the point I added the dynamic form widget.
<?php
// This function will run within each post array including multi-dimensional arrays
function ExtendedAddslash(&$params)
{
foreach ($params as &$var) {
// check if $var is an array. If yes, it will start another ExtendedAddslash() function to loop to each key inside.
is_array($var) ? ExtendedAddslash($var) : $var=addslashes($var);
unset($var);
}
}
// Initialize ExtendedAddslash() function for every $_POST variable
ExtendedAddslash($_POST);
$submission_id = $_POST['submission_id'];
$formID =$_POST['formID'];
$ip =$_POST['ip'];
$fname =$_POST['fname'];
$lname =$_POST['lname'];
$spousename =$_POST['spousename'];
$address =$_POST['address'][0]." ".$_POST['address'][1]." ".$_POST['address'][2]." ".$_POST['address'][3]." ".$_POST['address'][4]." ".$_POST['address'][5];
...
$db_host = 'localhost';
$db_username = 'xxxxx';
$db_password = 'xxxxx';
$db_name = 'xxxxx';
mysql_connect( $db_host, $db_username, $db_password) or die(mysql_error());
mysql_select_db($db_name);
// search submission ID
$query = "SELECT * FROM `tableName` WHERE `submission_id` = '$submission_id'";
$sqlsearch = mysql_query($query);
$resultcount = mysql_numrows($sqlsearch);
if ($resultcount > 0) {
mysql_query("UPDATE `tableName` SET
`fname` = '$fname',
`lname` = '$lname',
`spousename` = '$spousename',
`address` = '$address',
WHERE `submission_id` = '$submission_id'")
or die(mysql_error());
} else {
mysql_query("INSERT INTO `tableName` (submission_id, formID, IP, fname, lname, spousename, address)
VALUES ('$submission_id', '$formID', '$ip', '$fname', '$lname', '$spousename', '$address') ")
or die(mysql_error());
}
?>
It has been suggested that I explore using the PHP Explode() function. It's possible that may work, but can't get my head around how to apply the function to the PETLIST Array. Looking for suggestions or direction to find a solution for this challenge. Looking forward to any replies. I need help with search form. When someone enter in search form "dota allstars", if in mysql is no such a thing like that at table "news", i wonna to find similar data, how can i do that? I'm trying to do something that I thought was very simple about 2 weeks ago :-( I want to put a form on my site and link it to a database so when a user types a surname into the form they can search the db and the page will only display the surnames that match the search criteria. I got the db set up using phpmyadmin in about 2 minutes flat, but keep getting error messages when I write the php. I'm currently working with the below script, which keeps giving me the error 'unexpected T_string' on line 176 I've searched every forum and help site I can find, and the mysql and php manuals are just mind-boggling. I'm sure I'm making some really amateur mistake, but I'd really appreciate help with this! thanks all! <html> <head> <title>SEARCH RECORDS</title> </head> <body> <FORM NAME ="Search" METHOD ="POST" ACTION = "test3"> <INPUT TYPE = "TEXT" VALUE ="surname" NAME = "surname"> <INPUT TYPE = "Submit" Name = "Search" VALUE = "Search"> </FORM> </body> </html> <? $user_name = "*****"; $password = "*****"; $database = "*****"; $server = "localhost"; $db_handle = mysql_connect($server, $user_name, $password); $db_found = mysql_select_db($database, $db_handle); if ($db_found) { $SQL = "SELECT * FROM *****" WHERE surname=$_POST['surname']; $result = mysql_query($SQL); while ($db_field = mysql_fetch_assoc($result)) { print $db_field['Grave'] . "<BR>"; print $db_field['Surname'] . "<BR>"; print $db_field['Forenames'] . "<BR>"; print $db_field['Death_Date'] . "<BR>"; print $db_field['Birth_Year'] . "<BR>"; } mysql_close($db_handle); } else { print "Database NOT Found "; mysql_close($db_handle); } ?> Hi everyone, First off thank you very kindly to anyone who can provide some enlightenment to this total php/mysql newb. I am creating my first database and while I have been able to connect to the database within my php script (or so I believe), the form data does not appear to be actually getting sent. I have development experience in other languages but this is completely new to me so if I've missed something that appears painfully obvious like a parse error of some sort, I do apologize. I am creating a website using Godaddy as the hosting account, and attempting to connect to the mysql database at the following URL (maybe this is where I'm going wrong): "pnmailinglist.db.4662743.hostedresource.com" Below is my very simple code: <?php //Verify successful connection to database. $connect = mysql_connect("pnmailinglist.db.4662743.hostedresource.com", "*********", "*********"); if(!$connect) {die("Could not connect!"); } //Initialize variables with form data. $firstname = $_POST['FirstName']; $lastname = $_POST['LastName']; $email = $_POST['Email']; $howfound = $_POST['HowFound']; //Post data to database, or return error. mysql_select_db("pnmailinglist", $connect); mysql_query("INSERT INTO mailinglist (First, Last, Email, How_Found) VALUES ($firstname,$lastname,$email,$howfound)"); mysql_close($connect); echo "Thank you for joining our mailing list! We will contact you soon with the dates and times of our upcoming events."; ?> Thank you again very much for any pointers or hints as to where I'm screwing up. I get no runtime errors, no syntax errors, and the echo message does display fine at the end -- just no data when I go to check my database! Best Regards, CL Hello everyone,
I have a card database and everything works perfectly besides one thing..
I can't store
'in my database via my form, i however do can store them into the database via PHPMyAdmin. I dont know what i've been doing wrong and it really bothers me. If any of you guys could help me out. Here's all the code you would need to find the issue. This is the form file
<!doctype html>
<html>
<head>
<meta charset="utf-8">
<title>Edit Page</title>
</head>
<body>
<h1 align="center"> Add Cards</h1>
<form action="insert.php" method="POST">
<input type="text" name="name" placeholder="Name" />
<input type="text" name="color" placeholder="Color" />
<input type="text" name="type" placeholder="Type" />
<input type="text" name="subtype" placeholder="Sub Type" />
<input type="text" name="power" placeholder="Power" />
<input type="text" name="toughness" placeholder="Toughness" />
<br>
<input type="text" name="manacost" placeholder="Converted Mana Cost" />
<input type="text" name="rarity" placeholder="Rarity" />
<input type="text" name="expansion" placeholder="Expansion" />
<input type="text" name="foil" placeholder="Foil" />
<input type="text" name="stock" placeholder="Stock" />
<input type="submit" value="Save" />
</form>
</body>
</html>
This inserts it into my database.
<?php
($GLOBALS["___mysqli_ston"] = mysqli_connect("", "", "", , ))or die("cannot connect");
((bool)mysqli_query($GLOBALS["___mysqli_ston"], "USE e_industries"))or die("cannot select DB");
$name = $_POST['name'];
$color = $_POST['color'];
$type = $_POST['type'];
$subtype = $_POST['subtype'];
$power = $_POST['power'];
$toughness = $_POST['toughness'];
$manacost = $_POST['manacost'];
$rarity = $_POST['rarity'];
$expansion = $_POST['expansion'];
$foil = $_POST['foil'];
$stock = $_POST['stock'];
$sql="INSERT INTO Osiris (Name, Color, Type, Subtype, Power, Toughness, Manacost, Rarity, Expansion, Foil, Stock)
VALUES ('$name', '$color', '$type', '$subtype', '$power', '$toughness', '$manacost', '$rarity', '$expansion', '$foil', '$stock')";
$result=mysqli_query($GLOBALS["___mysqli_ston"], $sql);
if($result){
echo "Successful";
echo "<BR>";
echo "<a href='add.html'>Back to main page</a>";
}
else {
echo "ERROR";
}
?>
If anyone could help me out that would be great!
Edited by OsirisElKeleni, 05 October 2014 - 12:29 PM. I have deleted mysql info for safety reasons. Here are the two webpage's codes i'm using right now menu.php <? session_start(); if(!session_is_registered(myusername)){ header("location:login.php"); } ?> <html><title>ChronoServe - Saving Time</title> <link href="style.css" rel="stylesheet" type="text/css"> <body> <table width="100%" border="0" cellpadding="0" cellspacing="0" class="container"> <tr> <td> <table width="335px" height="50%" border="1" align="center" cellpadding="0" cellspacing="0" class="centered"> <tr> <td> <form method="post" action="insertvalues.php"> <table width="100%" border="0" align="center" cellpadding="3" cellspacing="10"> <tr> <td colspan="2"><div align="center" class="font2">Activation Information</div></td> </tr> <tr> <td colspan="2"></td> </tr> <tr> <td width="40%" class="font3">First Name :</td> <td width="60%"> <div align="center"> <input name="firstname" type="text" class="font3" id="firstname" maxlength="25" /> </div></td> </tr> <tr> <td class="font3">Last Name :</td> <td> <div align="center"> <input name="lastname" type="text" class="font3" id="lastname" maxlength="25" /> </div></td> </tr> <tr> <td height="28" class="font3">Phone Number :</td> <td> <div align="center"> <input name="pnumber" type="text" class="font3" id="pnumber" maxlength="10" /> </div></td> </tr> <tr> <td class="font3">Personnel Activated :</td> <td> <div align="center"> <input name="numberactivated" type="text" class="font3" id="numberactivated" maxlength="3" /> </div></td> </tr> <tr> <td height="37" colspan="2"></td> </tr> <tr> <td colspan="2"><div align="center"> <input name="submit" type="Submit" class="font3" value="Submit" /> </div> </td> </tr> </table> </form></td> </tr> </table> </td> </tr> </table> </body> </html> insertvalues.php <?php if(isset($_POST['Submit'])) { $firstname = $_POST['firstname']; $lastname = $_POST['lastname']; $pnumber = $_POST['pnumber']; $numberactivated = $_POST['numberactivated']; mysql_connect ("deleted", "deleted", "deleted") or die ('Error: ' . mysql_error()); mysql_select_db ("deleted"); $query = "INSERT INTO disney_database (id, firstname, lastname, pnumber, numberactivated, date) VALUES ('NULL', '".$firstname."', '".$lastname."', '".$pnumber."', '".$numberactivated."', 'NULL')"; mysql_query($query) or die('Error updating database'); header("location:menu.php"); echo "Database Updated With: ".$firstname."" - "".$lastname."" - "".$pnumber."" - "".$numberactivated.""; } else { echo "Database Error" { ?> Here is my problem. I set a one <form> on every form field I have including the submit button. Now whenever I press the submit button it redirects to insertvalues.php which it should be doing. In insertvalues i told it to query the form data and post it into my database's table. Its not doing that and tells me that it has a database error which i set it to tell me if something goes wrong. Anyone can help me? BTW I can manually query in the information using sql with phpmyadmin. so can someone please review my code for me? thanks big help! You can see what is happening. Visit www.chronoserve.com The username and password are "admin" I am having difficulty inserting the following codes values into my database. I know that the variables contain a value as I am displaying them on the output screen. Code: [Select] $link = mysql_connect($db_host,$db_user,$db_pass) or die('Unable to establish a DB connection'); mysql_select_db($db_database,$link); mysql_query("SET names UTF8"); $usr = $userid; $golfer = $_REQUEST['golfer']; $tourney = $_REQUEST['tournament']; $backup = $_REQUEST['backup']; date_default_timezone_set('US/Eastern'); $time = date("Y-m-d H:i:s"); $t_id=1; mysql_query("INSERT INTO weekly_picks (t_id, tournament, user, player, backup, timestamp) VALUES ('$t_id', '$tournament', '$usr', '$golfer', '$backup', '$time') or die('Error, insert query failed')"); echo $t_id; echo "<br />"; echo $tourney; echo "<br />"; echo $usr; echo "<br />"; echo $golfer; echo "<br />"; echo $backup; echo "<br />"; echo $time; echo "<br />"; echo $userdetail['email']; echo "<br />"; $to = $userdetail['email']; $subject = "Weekly Tournament Pick for: $tourney"; $message = "This email is to confirm your pick for $tourney has been received. Your pick is: $golfer. Your backup pick is: $backup The time it was submitted was $time Do not reply to this email, the mailbox does not exist. Contact me with any issues at xxxxxxx@xxxxxxx.com"; $from = "noreply@chubstersgcc.com"; $headers = "From:" . $from; mail($to,$subject,$message,$headers); ?> <div> <p>This is to confirm that your pick has been submitted for the following tournament: <?php echo $tourney; ?>. A confirmation of your pick as also been emailed to you.</p> <p>Your golfer: <?php echo $golfer; ?></p> <p>Your backup: <?php echo $backup; ?></p> <p>Your pick was submitted at: <?php echo $time; ?></p> </div> |
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