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PHP - Moved: How To Get A Php Search Form To Suggest Alternatives In Mysql Database?
This topic has been moved to Ajax Help.
http://www.phpfreaks.com/forums/index.php?topic=326703.0 Similar TutorialsI'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... This topic has been moved to MySQL Help. http://www.phpfreaks.com/forums/index.php?topic=334042.0 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! 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...
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 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. 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". 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 This topic has been moved to Miscellaneous. http://www.phpfreaks.com/forums/index.php?topic=326179.0 Hey guys, I have the following form (attached to this post as a Thumbnail).
I want to make it user-friendly, specifically, bring it up to the modern web standards, namely, I am looking for things that that I think are modern and are user friendly and are great when considered by themselves, but together they may even conflict. And I need some help figuring out what will work best for my situation and what is technically possible.
Right now the user enteres fields into the form, into the editable fields on the left. Then user can save the fields by clicking Save, or click Del to delete the row. User can Add a new row at the bottom.
The things I think will improve the form a * implement a double-click feature on the editable fields to make them editable, and save them automatically upon `blur` event * user can use `Tab` key to navigate from one `<input>` field to the next, normally, as before, and edit values inside. * user can save all fields entered so far, and restore them via some mechanism (i.e. either to web page or to clipboard, and restore via direct paste for example, or file upload..) * Maybe I can use DataTables, but I have never used it so I don't know what could be done Can you help suggest form or UI improvements for the current form that I have? I may also have some questions on first steps or first examples of implementing particular ones (possibly making a separate post for those) Attached Files form.png 7.94KB 0 downloads Now i use a very complex normal for, how i can use a foreach? This topic has been moved to Third Party PHP Scripts. http://www.phpfreaks.com/forums/index.php?topic=321218.0 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 This topic has been moved to Miscellaneous. http://www.phpfreaks.com/forums/index.php?topic=343257.0 This topic has been moved to MySQL Help. http://www.phpfreaks.com/forums/index.php?topic=343372.0 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? This topic has been moved to Other Libraries and Frameworks. http://www.phpfreaks.com/forums/index.php?topic=359217.0 This topic has been moved to MySQL Help. http://www.phpfreaks.com/forums/index.php?topic=308347.0 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); } ?> |
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