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PHP - Execute Sql Query By Clicking Link/button
So I am trying to execute an sql query by clicking a button or a link. Ultimately I want users to be able to click the "Fav" button and it then fades into "Added to favorites!" (I am building a favorite system which first collects data on page load like userid and itemid and then it needs to send it to the DB after a button is clicked)
The fade and stuff is not the most important part now (although a tip on doing that would be awesome) but the most important part is how to get this functional: PHP Code: <?php // Start Favorite System! $user =& JFactory::getUser(); $userid = $user->id; if ($userid == 62) { echo "userID: ".$userid."<p />"; if ($user->guest) {echo "You are a guest<p />";} $itemid = $this->item->id; echo "itemID: ".$itemid."<p />"; $catid= $this->item->category->id; echo "catID: ".$catid."<p />"; $query = "INSERT INTO jos_k2_fav_xref (userID, itemID, catID) VALUES ('$userid', '$itemid', '$catid')"; // Need a way to make this into a button $run = mysql_query($query) or die(mysql_error()); } else {} ?> Currently all the values get stored in the DB as the page is loaded. I however want to be able to click on something to execute that query how do i do that? I read a bunch of stuff on javascript and ajax but it got me nowhere... Help is greatly appreciated! Similar TutorialsI have this piece of code: Code: [Select] if (isset($_POST['type1'])) { mysql_query("UPDATE tabke SET status = 1 WHERE id = $rid", $c2) or die(mysql_error()); } if (isset($_POST['type2'])) { mysql_query("UPDATE table SET status = 1 WHERE id = $rid", $c2) or die(mysql_error()); } if (isset($_POST['type3'])) { mysql_query("UPDATE table SET status = 1 WHERE id = $rid", $c2) or die(mysql_error()); } ?> <br /> <form method="post" action=""> <input type="hidden" name="pageid" value="plyrmgmt"> <input type="hidden" name="action" value="changeBan"> <input type="hidden" name="uid" value="<?php echo $hr_uid; ?>"> <input type="hidden" name="repid" value="<?php echo $rid; ?>"> <input type="submit" name="type1" value="Choice1" onClick="this.form.submit()"> </form> <form method="post" action=""> <input type="hidden" name="pageid" value="plyrmgmt"> <input type="hidden" name="action" value="changeMute"> <input type="hidden" name="uid" value="<?php echo $hr_uid; ?>"> <input type="hidden" name="repid" value="<?php echo $rid; ?>"> <input type="submit" name="type2" value="Choice2" onClick="this.form.submit()"> </form> <form method="post" action=""> <input type="hidden" name="pageid" value="plyrmgmt"> <input type="hidden" name="action" value="changeLock"> <input type="hidden" name="uid" value="<?php echo $hr_uid; ?>"> <input type="hidden" name="repid" value="<?php echo $rid; ?>"> <input type="submit" name="type3" value="Choice3" onClick="this.form.submit()"> </form> Now, for some reason when I click on Choice1, the query doesn't execute. Also, when I click on Choice2 or Choice3, the URL in my browser doesn't change for some odd reason...it stays the same as it was prior to clicking the Submit Button. Can anyone point out some errors I have? Thanks, Mark. Hi, Im starting to write a simple (so I thought! ) project at work, I have no problems with the HTML, CSS or even the php mysql setup, but I am struggling with a simple piece of code! I have a list box, that populates from a table column using the following code: Code: [Select] <?php function select(){ $query="SELECT appname FROM kpe_apps"; $result = mysql_query($query); echo '<select name="item" onchange="this.form.submit()">'; while($nt=mysql_fetch_array($result)){ echo '<option value="' . $nt['id'] . '">' . $nt['appname'] . '</option>'; } echo '</select>'; } ?> I then called the function on the page I needed. which works fine, now all I need is whatever is selected in the listbox other columns in the same table relating to the selected item are seen. I just cant seem to do it, loads of people are using jquery and other code, surely this can be done in php? any help would really be appreciated.. Many Thanks MOD EDIT: code tags fixed, linefeeds and indenting added. Hi. I have a query that determines if user is online/offline. Works perfectly when page loads. I want to change to auto run query every X seconds. NOTES: - The query is on a .php page with multiple other queries. -IF it is possible to still run the query with-in this page? or make a separate file and include in an iframe? i would like to avoid the iframe and just $output the result of query every X seconds. Here is my $query code: $query = "SELECT `manager_id` FROM #__profiles_xref WHERE profileid = '$profid'"; $onofflineid=doSelectSql($query,1); foreach ($onofflineid as $isonoffline){ $profmanagerid=$isonoffline->manager_id; } $query = "SELECT `session_id` FROM table_session WHERE userid = '$propmanagerid'"; $onofflinestatus=doSelectSql($query,1); if (count($onofflinestatus)>0) { $output['PROPSTATUS']='Online'; } else { $output['PROPSTATUS']='Offline'; } Unable to execute query (SELECT * FROM lb-players) in the database : You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near '-players' at line 1 I don't know why i am getting this error. All i can think of that is taking away lb from "lb-players" when you add a " - "? Code: [Select] [m]<?php $conn = mysql_connect("23.23.23.23", "unknown", "itsAsecertxD");//ignore this xD if (!$conn) { echo "Unable to connect to the database : " . mysql_error(); exit; } if (!mysql_select_db("minecraft-rc")) { echo "Unable to select database mydbname : " . mysql_error(); exit; } $sql = 'SELECT * FROM lb-players'; $result = mysql_query($sql); if (!$result) { echo "Unable to execute query ($sql) in the database : " . mysql_error(); exit; } if (mysql_num_rows($result) == 0) { echo "No rows found, nothing to display."; exit; } while ($row = mysql_fetch_assoc($result)) { echo $row["playername"]; } mysql_free_result($result); ?>[/m] tell me if i posted this in the wrong place its been forever since i last been on this site :$ Hi all ! The following code has two similar queries. Query 1 and Query 2. While Query 1 works just fine, Query 2 simply fails persistently. I have not been able to figure out why?
I have checked every single variable and field names but found no discrepancy of any sort or any mismatch of spelling with the fields in the database.
I would be grateful if anybody can point out what is preventing the second query from returning a valid mysqli object as the first one does. Here is the code for the two.
<?php
$db_host = "localhost";
$db_user ="root";
$db_pass ="";
$db_database ="allscores";
//////////////// CHECKING VARIABLE FILEDS IN A UPDATE QUERY //////////////////////////////////
//////////////// QUERY 1 /////////////////////////////////////////////////////////////////////
/*
$db_database ="test";
$db_table ="users";
$RecNo = 1;
$field1 = 'name';
$field2 = 'password';
$field3 = 'email';
$field4 = 'id';
$val1 = "Ajay";
$val2 = "howzatt";
$val3 = "me@mymail.com";
$con = mysqli_connect($db_host,$db_user,$db_pass,$db_database) or die('Unable to establish a DB connection');
$query = "UPDATE $db_table SET $field1 = ?,
$field2 = ?, $field3 = ?
WHERE $field4 = ? "; // session terminated by setting the sessionstatus as 1
$stmt = $con->prepare($query);
var_dump($stmt);
$stmt->bind_param('sssi',$val1,$val2,$val3,$RecNo);
if($stmt->execute())
{
$count = $stmt->affected_rows;
echo $count;
}
//////////////// QUERY 1 END /////////////////////////////////////////////////////////////////////
*/
//////////////// QUERY 2 /////////////////////////////////////////////////////////////////////
$con = mysqli_connect($db_host,$db_user,$db_pass,$db_database) or die('Unable to establish a DB connection');
$table = 'scores';
$date = date('Y-m-d H:i:s');
/*
$prestr = "Mrt_M_";
$STATUS = $prestr."Status";
$S_R1 = $prestr."Scr_1";
$S_R2 = $prestr."Scr_2";
$PCT = $prestr."PPT";
$DPM = $prestr."DSP";
$TIMETAKEN = $prestr."TimeTaken";
*/
$STATUS = "Mrt_M_Status";
$S_R1 = "Mrt_M_Scr_1";
$S_R2 = "Mrt_M_Scr_2";
$PPT = "Mrt_M_PPT";
$DSP = "Mrt_M_DSP";
$TIMETAKEN = "Mrt_M_TimeTaken";
$TimeOfLogin = $date;
$no_of_logins = 10;
$time_of_nLogin = $date;
$m_scr_row1 = 5;
$m_scr_row2 = 5;
$m_ppt = 20;
$m_dsp = 60;
$m_time = 120;
$date = $date;
$RecNo = 24;
$query = "UPDATE $table SET
TimeOfLogin = ?,
no_of_logins = ?,
time_of_nLogin = ?,
$S_R1 = ?,
$S_R2 = ?,
$PPT = ?,
$DSP = ?,
$TIMETAKEN = ?,
$STATUS = '1',
TimeOfLogout = ?,
WHERE RecNo = ?";
$stmt = $con->prepare($query);
var_dump($stmt);
$stmt->bind_param('sisiiddssi',$TimeOfLogin,$no_of_logins,$time_of_nLogin,$m_scr_row1
$m_scr_row2,$m_ppt,$m_dsp,$m_time,$date,$RecNo);
if($stmt->execute()) echo " DONE !";
?>
Thanks to all
Im displaying the results of a database by using a while loop... so Im trying to figure out how to delete some entries from the datbase. What I wanted to do is add a button in each result that will have the value "delete"... but I cant figure out how to do this. Im kinda new to mysql. I just want it so when I click the button the result will get deleted from the database. I did some research on this but I cant really grasp the examples. Can anybody help me out with this one? Thanks! I have no clue about to go abt this one: I have form which has some: Code: [Select] <table> <tr> <td> <form> Title display info </form> </td> </tr> </table> I need to add an additional link next to title such that when someone clicks on the link the table expands and displays rsult of a query. Code: [Select] <table> <tr> <td> <form> Title LINK if( link is clicked ) { $query=select display query results } else { dont display query results display info } </form> </td> </tr> </table> So, I just don't know where to post this question. I've written a quiz program where 50 English vocab words are pulled from a dB and displayed on one page with the order scrambled. My students then type the Spanish translation for each. When they submit the form the next page grades their work. They have to get a 90 or better for the grade to be automatically recorded, so I want them to click the back button and go back and make corrections. The problem is when they use a Chrome browser and click the back button the program re-sorts the order of questions. It leaves the order of their answers the same. So what they answered for number 1 is still in the field for number 1, but the questions (the English vocab word) is different. Chrome does this consistently so I thought this was a Chrome issue only. Then one of my students said the same thing happened to her when using Safari. So here is my question, how can I stop Chrome (or any other browser) from re-running my server-side PHP scripts (re-sorting the vocab array) when clicking the back button?
Here is the code for students taking the quiz:
<form method="post" action="portal.php?load=vquiz_grade" style="margin-left: 50px; line-height: 30px;">
$max = $count-1;
?>
//print "<br>";
<input type="hidden" name="max" value="<?print $max?>"> And here is the code showing them what they missed.
<?
// Total grade
print "You missed ".$wcount." words.<br>";
print "<table>";
// Grade response and set color
if ($wrong == 1){ if ($grade >= 90){
$query = "SELECT * FROM inno_vocab_assignments WHERE key_code = '$key_code' AND uname = '$secure_uname' AND teacher = '$teacher'";
// Prevent student from changing vocab list so as to do an easier list
// Insert grade
}
Thank you!
This topic has been moved to Third Party PHP Scripts. http://www.phpfreaks.com/forums/index.php?topic=355955.0 Seems to be a basic question, but I couldn't find an answer nor figure it out on my own. Basically I have a script that takes out specific data out of the database, the script works on its own, now I just need a way to make the user execute it with a link or a button. Example: Category: [Smileys] - [Category2] - [Category3] - etc. As soon as the user clicks on [Smileys] all data in the database which contains the word smileys in the category field gets selected and outputted as a list. Again the script works, I just need to be able to execute it with a button. If I understood it correctly I have to run the script by adding a if (isset($_POST['Smileys'])) { in front of the script. But how do I build the connection with the text link? I have a PHP application where I am trying to execute a PHP routine when the user clicks on a link to go to another page. The routine needs to execute before I use a header location command to send the user to the next location. I can't do this all with JavaScript, but it could be part of the solution if necessary. I'd prefer to just use PHP if I could, but I'm not sure that this is possible. Can anyone share with me a brief example of how I could do this? Thanks! Hi
I'm trying to change an image when a certain link is clicked. The image is a logo of a company and the links that will be clicked is english, arabic, and Chinese.
So if someone clicks on Chinese then the logo changes to the Chinese version of the normal logo.
Here is my html
<?php
$langArabic='';
$langChinese='';
$langSpanish='';
$langEng='';
$linkArabic=getSEOLink($page_id);
$linkChinese=getSEOLink($page_id);
$linkSpanish=getSEOLink($page_id);
$linkEng=getSEOLink($page_id);
$queryLink = mysql_query("SELECT * FROM $_SEO_TABLE WHERE `id_page`='".$page_id."'");
if(mysql_num_rows($queryLink) > 0){
$resultLink = mysql_fetch_object($queryLink);
if($resultLink->url != ''){
$linkEng = $_HTTP_ADDRESS."".$resultLink->url;
if($page_id == 1 || $page_id == '1'){
$linkEng = $_HTTP_ADDRESS;
}
}
if($resultLink->url_arabic != ''){
$linkArabic = $_HTTP_ADDRESS."".$resultLink->url_arabic;
}
if($resultLink->url_chinese != ''){
$linkChinese = $_HTTP_ADDRESS."".$resultLink->url_chinese;
}
if($resultLink->url_spanish != ''){
$linkSpanish = $_HTTP_ADDRESS."".$resultLink->url_spanish;
}
}
switch (urlExtension()) {
case 'html':
$langClass = '';
$langEng='langActive';
$langTag='en';
break;
case 'ar':
if($page->title_arabic == ''){
$langClass = '_arabic';
}else{
$langClass = '_arabic';
}
$langArabic='langActive';
$langTag='ar';
break;
case 'es':
$langClass = '';
$langSpanish='langActive';
$langTag='es';
break;
case 'cn':
$langClass = '';
$langChinese='langActive';
$langTag='cn';
break;
default:
$langClass = '';
$langEng='langActive';
$langTag='en';
}
?>
<a class="main-logo<?php echo $langClass;?>" href="javascript:void(0);"><img src="<?php echo $_HTTP_ADDRESS;?>images/logo-letters.png" /></a>
<div class="language-selection-wrapper<?php echo $langClass;?> overlay-bg<?php echo $langClass;?> border-bottom-white<?php echo $langClass;?> padding-level-one<?php echo $langClass;?>">
<div class="language-inner-wrapper<?php echo $langClass;?>">
<a class="<?php echo $langEng;?>" href="<?php echo $linkEng;?>">English</a><span>|</span>
<a class="<?php echo $langArabic;?>" href="<?php echo $linkArabic;?>">العربيّة</a><span>|</span>
<a class="<?php echo $langSpanish;?>" href="<?php echo $linkSpanish;?>">Español</a><span>|</span>
<a class="<?php echo $langChinese;?>" href="<?php echo $linkChinese;?>">中文</a>
<div class="clearboth"></div>
</div>
I have a php form in which ,on click of a edit button the contents of that table row will be displayed in a text area (which appears on the click of that edit button).Below the textarear i have a submit and delete button,i can perform any action.Ater performing the save or delete click the text area should be hidden. this is not happening,tried with js also. pasting the code below Code: [Select] <form name='add_title' method='post' action=''><?php echo "<br><br>"; if(isset($_POST['update'])) { if(isset($_GET['id'])) { $id = $_GET['id']; $comments=$_POST['comment']; $query1 = "update comments set comments='$comments' WHERE id = '$id' "; $result = mysql_query($query1); //$row=mysql_fetch_row($result); } } if(isset($_POST['delete'])) { if(isset($_GET['id'])) { $id = $_GET['id']; $comments=$_POST['comment']; $query1 = "delete from comments WHERE id = '$id' "; $result = mysql_query($query1); } } $query="select count(*) from comments"; $result =mysql_query($query); $row=mysql_fetch_row($result); $count= $row[0]; if($count!=0) { $query="select *from comments order by id desc"; $result=mysql_query($query); echo "<table width='700' align='center' border=1><tr bgcolor='green' align='center'><td><font color='black'> SlNo </font></td><td><font color='black'>Comment</font></td><td><font color='black'> Date Posted </font></td><td>Edit/Delete</td></tr></b>"; while($row = mysql_fetch_row($result)) { $d=$row['0']; $id=$row['0']; $name=$row['1']; $date_uploaded=$row['3']; echo "<td>$id</td><td>$name</td>". "<td>$date_uploaded</td>". "<td><a href='edit_comments.php?id=$id'>Edit </a></td>". "</tr>"; } if(isset($_GET['id'])) { $id = $_GET['id']; $query = "SELECT comments FROM comments WHERE id = '$id' "; $result = mysql_query($query); $row=mysql_fetch_row($result); echo"<center><table width='700' border='1'><tr><td > <textarea name='comment' cols='80' rows='10'>$row[0]</textarea> <td></tr><tr><td><a href='edit_comments.php?id=$id'><input type='submit' name='update' id='update' value='Save'></a><a href='edit_comments.php?id=$id'><input type='submit' name='delete' id='delete' value='Delete'></a> </td><td></td></center>"; } echo "</table>"; } else { echo "<table width='700' align='center'><tr bgcolor='white' align='center'><td><font color='red' size=2>No comments for you.</font></td></tr></table>"; } ?> Edit: Please use [code][/code] tags when posting code. Hi i am working one of my assignment for school.. i just studied php.. my problem is, is it possible to load another form once i clicked a radio button? if its a yes... can you teach me how to do that? thank you! Folks,
Look at this weird thing. I load the page and get echoed as expected: Did Not REQUEST_METHOD! Then, I click the SUBMIT button and to my astonishment I get echoed: Did Not POST->Submit! Got 2 buttons. Same result whenever clicking any. Why is that ? Check it out:
<?php
//include('error_reporting.php');
ini_set('error_reporting',E_ALL);//Same as: error_reporting(E_ALL);
ini_set('display_errors','1');
ini_set('display_startup_errors','1');
?>
<form name = "submit" method = "POST" action="">
<label for="domain">Domain:</label>
<input type="text" name="domain" id="domain" placeholder="Input Domain">
<br>
<label for="domain_email">Domain Email:</label>
<input type="email" name="domain_email" id="domain_email" placeholder="Input Domain Email">
<br>
<label for="url">Url:</label>
<input type="url" name="url" id="url" placeholder="Input Url">
<br>
<label for="link_anchor_text">Link Anchor Text:</label>
<input type="text" name="link_anchor_text" id="link_anchor_text" placeholder="Input Link Anchor Text">
<br>
<textarea rows="10" cols="20">Page Description</textarea>
<br>
<label for="keywords">Keywords:</label>
<input type="text" name="keywords" id="keywords" placeholder="Input Keywords related to Page">
<br>
<input type="checkbox" name="alert_visitor_type" id="alert_visitor_type" value="Give Alert: Visitor Type">
</label for="alert_visitor_type">Give Alert: Visitor Type</lablel>
<input type="checkbox" name="alert_visitor_potential" id="alert_visitor_potential" value="Give Alert: Potential Visitor">
</label for="alert_visitor_potential">Give Alert: Potential Visitor</lablel>
<br>
<input type="radio" name="tos_agree" id="tos_agree_yes" value="yes">
<label for="tos_agree_yes">Yes:</label>
<input type="radio" name="tos_agree" id="tos_agree_no" value="no">
<label for="tos_agree_no">No:</label>
<br>
<label for="tos_agreement">Agree to TOS or not ?</label>
<select name="tos_agreement" id="tos_agreement">
<option value="yes">Yes</option>
<option value="no">No</option>
</select>
<br>
<button type="submit" value="submit">Submit</button><br>
<button type="submit">Submit</button>
<br>
<input type="reset">
<br>
</form>
<?php
if($_SERVER['REQUEST_METHOD'] === 'POST')
{
if(isset($_POST['submit']))
{
mysqli_report(MYSQLI_REPORT_ALL|MYSQLI_REPORT_STRICT);
mysqli_connect("localhost","root","","test");
$conn->set_charset("utf8mb4");
$query = "INSERT into links (domain,domain_email,url,link_anchor_text,page_description,keywords,alert_visitor_type,alert_visitor_potential) VALUES (?,?,?,?,?,?,?,?)";
$stmt = mysqli_stmt_init($conn);
if(mysqli_stmt_prepare($stmt,$query))
{
mysqli_stmt_bind_param($stmt,'ssssssss',$_POST['domain'],$_POST['domain_email'],$_POST['url'],$_POST['link_anchor_text'],$_POST['page_description'],$_POST['keywords'],$_POST['alert_visitor_type'],$_POST['alert_visitor_potential']);
if(mysqli_stmt_execute($stmt) === FALSE)
{
die("Error\" . mysqli_stmt_error()");
}
mysqli_stmt_close($stmt);
mysqli_close($conn);
}
else
{
die("Did not INSERT!");
}
}
else
{
die("Did Not POST->Submit!");
}
}
else
{
die("Did Not REQUEST_METHOD!");
}
?>
Dear All I have a form say i have selected Volva from drop down box . So without clicking on submit button when i move to next field that is `<input>` the value that is volva should get store in php variable <html> <body> <form action=""> <select name="cars"> <option value="volvo">Volvo</option> <option value="saab">Saab</option> <option value="fiat">Fiat</option> <option value="audi">Audi</option> </select> <input type="text" name="abc" /> </form> </body> </html> Hi.. I have an issue on my project.I want to open a particular page of a PDF file by clicking a hyper link.How should I do it?any idea ? I used the following code <?php header("Content-Type: application/pdf"); $pdfFile="readme.pdf#page=10"; ?> <html> <head> <title>Untitled Document</title> </head> <body> <a href="http://<?php echo $pdfFile; ?>">Click Here</a> </body> </html> But it showing an alert like following File does not begin with '%PDF-' After clicking alert showing a black screen.. can u please send me the entire code? Thanks... This topic has been moved to JavaScript Help. http://www.phpfreaks.com/forums/index.php?topic=357107.0 This topic has been moved to Ajax Help. http://www.phpfreaks.com/forums/index.php?topic=326913.0 Goal: To have a gallery that downloads images from the folder I previously uploaded to in a previous script. Bug: When I load the page the thumbnail comes up as broken and when I click on the thumbnail to get the bigger picture it comes up with the following error message: "Firefox doesn't know how to open this address, because the protocol (c) isn't associated with any program." <?php include 'db.inc.php'; //connect to MySQL $db = mysql_connect(MYSQL_HOST, MYSQL_USER, MYSQL_PASSWORD) or die ('Unable to connect. Check your connection parameters.'); mysql_select_db(MYSQL_DB, $db) or die(mysql_error($db)); //change this path to match your images directory $dir ='C:/x/xampp/htdocs/images'; //change this path to match your thumbnail directory $thumbdir = $dir . '/thumbs'; ?> <html> <head> <title>Welcome to our Photo Gallery</title> <style type="text/css"> th { background-color: #999;} .odd_row { background-color: #EEE; } .even_row { background-color: #FFF; } </style> </head> <body> <p>Click on any image to see it full sized.</p> <table style="width:100%;"> <tr> <th>Image</th> <th>Caption</th> <th>Uploaded By</th> <th>Date Uploaded</th> </tr> <?php //get the thumbs $result = mysql_query('SELECT * FROM images') or die(mysql_error()); $odd = true; while ($rows = mysql_fetch_array($result)) { echo ($odd == true) ? '<tr class="odd_row">' : '<tr class="even_row">'; $odd = !$odd; extract($rows); echo '<td><a href="' . $dir . '/' . $image_id . '.jpg">'; echo '<img src="' . $thumbdir . '/' . $image_id . '.jpg">'; echo '</a></td>'; echo '<td>' . $image_caption . '</td>'; echo '<td>' . $image_username . '</td>'; echo '<td>' . $image_date . '</td>'; echo '</tr>'; } ?> </table> </body> </html> Any help appreciated. |
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