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PHP - Adding Something To Database With Check
Hey All, first off Merry X-Mas
ok so I'm trying to insert something into the batabse in one of three areas' let me explain there are three fields in the users database bounceone, bouncetwo and bouncethree now what I want to do is when the user clicks add to bounce list insert a value into the databse in one of these three hole, if bounceone is empty put it there, if not check bouncetwo if it's empty put it there, if not check the final one and if it's empty put it there else echo sorry please either remove a bounce any ideas on how this can be done. Similar TutorialsHello there, My current song request form process looks like: <?php if(isset($_POST['submit'])) { $to = 'address@website.com' ; $name = trim($_POST['name']); $message = trim($_POST['message']); $artist = trim($_POST['artist']); $song = trim($_POST['song']); $subject="Name:$name - Message:$message - Artist:$artist - Song:$song"; $headers = 'MIME-Version: 1.0' . "\r\n"; $headers .= 'Content-type: text/html; charset=iso-8859-1' . "\r\n"; $message = "<table> <tr><td>Name</td><td>".$_POST['name']."</td></tr> <tr><td>Message</td><td>".$_POST['message']."</td></tr> <tr><td>Artist</td><td>".$_POST['artist']."</td></tr> <tr><td>Song</td><td>".$_POST['song']."</td></tr> </tr></table>" ; mail($to, $subject, $message, $headers); header('Location: song-requests-success.php'); } ?> I'm currently using Javascript for blank validation, however, people seem to be getting through it somehow. I'm looking to implement PHP checkers that will output back to the request form with something like: Name field blank. Would this method work? $name = trim($_POST['name']); $artist = trim($_POST['artist']); $song = trim($_POST['song']); if(empty($name)){ $name1 .= "<br>Name is empty."; } if(empty($artist)){ $artist1 .= "<br>Artist is empty."; } if(empty($song)){ $song1 .= "<br>Song is empty."; } On form: <?php if(!empty($name)){ echo "".($name1).""; } ?> <?php if(!empty($artist)){ echo "".($artist1).""; } ?> <?php if(!empty($song)){ echo "".($song1).""; } ?> Would recalling the trim's be nessessary as they have already been specified for the email title. Would this need to be implemented into 'isset'. Many thanks. Hello i have a text box and a button on my webpage and want to get the text box to find the username i type into it from the database and when it does it will then go to number and delete any text that is under usernames number field as you can see i have no idea how. Code: [Select] $result = mysql_query("DELETE FROM mytable WHERE username='$username1' AND number='$number1'" ); Code: [Select] <form name="form1" method="post" action="p.php"> <p><input type="submit" value="Enter" /></p> <input name="username" type="text" id="username" size="15" maxlength="35" value="<?php echo $username1; ?>" /> </form> How do you check if a MySQL or MySQLi server is available through php. I'm trying to make a short script that makes sure the users server can run the application. Like all the popular ones do Hi guys, I need your help. I am checking on a database as I want to see if I have the same value in the url and in the database. Code: [Select] <?php session_start(); define('DB_HOST', 'localhost'); define('DB_USER', 'mydbuser'); define('DB_PASSWORD', 'mydbpass'); define('DB_DATABASE', 'mydbtable'); $errmsg_arr = array(); $errflag = false; $link = mysql_connect(DB_HOST, DB_USER, DB_PASSWORD); if(!$link) { die('Failed to connect to server: ' . mysql_error()); } $db = mysql_select_db(DB_DATABASE); if(!$db) { die("Unable to select database"); } function clean($var){ return mysql_real_escape_string(strip_tags($var)); } $username = clean($_GET['user']); $password = clean($_GET['pass']); $test = clean($_GET['test']); $public = clean($_GET['public']); if (isset($_GET['user']) && (isset($_GET['pass']))) { if($username == '' || $password == '') { $errmsg_arr[] = 'username or password are missing'; $errflag = true; } } elseif (isset($_GET['user']) || (isset($_GET['test'])) || (isset($_GET['public']))) { if($username == '' || $test == '' || $public == '') { $errmsg_arr[] = 'user or others are missing'; $errflag = true; } } if($errflag) { $_SESSION['ERRMSG_ARR'] = $errmsg_arr; echo implode('<br />',$errmsg_arr); } else { $qry="SELECT * FROM members WHERE username='$username' AND passwd='$password'"; $result=mysql_query($qry) or die('Error:<br />' . $qry . '<br />' . mysql_error()); if ($username && $password) { if(mysql_num_rows($result) > 0) { $qrytable1="SELECT images, id, test, links, Public FROM user_list WHERE username='$username'"; $result1=mysql_query($qrytable1) or die('Error:<br />' . $qry . '<br />' . mysql_error()); while ($row = mysql_fetch_array($result1)) { echo "<p id='test'>"; echo $row['test'] . "</p>"; echo '<p id="images"> <a href="images.php?test=test&id='.$row['id'].'">Images</a></td> | <a href="http://' . $row["links"] . '">Link</a> </td> | <a href="delete.php?test=test&id='.$row['id'].'">Delete</a> </td> | <span id="test">'.$row['Public'].'</td>'; } } else { echo "user not found"; } } elseif($username && $test && $public) { $qry="SELECT * FROM members WHERE username='$username'"; $result1=mysql_query($qry) or die('Error:<br />' . $qry . '<br />' . mysql_error()); if(mysql_num_rows($result1) > 0) { $qrytable1="SELECT Public FROM user_list WHERE username='$username' && test='$test'"; $result2=mysql_query($qrytable1) or die('Error:<br />' . $qry . '<br />' . mysql_error()); if(mysql_num_rows($result2) > 0) { $row = mysql_fetch_row($result2); mysql_query("UPDATE user_list SET Public=('$_GET[public]') WHERE username='$username' AND test='$test'"); echo "update!"; } else { echo "already updated!"; } } else { echo "user not found"; } } } ?> This is the function I use to check the value in the database: Code: [Select] if (mysql_affected_rows($result2) > 0) { mysql_query("UPDATE user_list SET Public=('$_GET[public]') WHERE username='$username' AND test='$test'"); echo "you have update it!"; } else if (mysql_affected_rows($result2) < 0) { echo "it is not on the database"; } else { echo "you have already updated!"; } When i input the different value in a url bar while the records are not the same as the value in the url and in the database, i can't get passed and I am keep getting "you have already updated!!" when the value in a database are different than I have input in a url. Do you know how i can get pass it when I have input the different value in the url while it is not the same in the database? Any advice would be much appreicated. Thanks, Mark Sorry going to be a big post, but trying to query a sql database for a value Original code <?php if($this->item->params->get('catItemExtraFields') && count($this->item->extra_fields)): ?> <!-- Item extra fields --> <div class="catItemExtraFields"> <h4><?php echo JText::_('K2_ADDITIONAL_INFO'); ?></h4> <ul> <?php foreach ($this->item->extra_fields as $key=>$extraField): ?> <?php if($extraField->value): ?> <li class="<?php echo ($key%2) ? "odd" : "even"; ?> type<?php echo ucfirst($extraField->type); ?> group<?php echo $extraField->group; ?>"> <span class="catItemExtraFieldsLabel"><?php echo $extraField->name; ?></span> <span class="catItemExtraFieldsValue"><?php echo $extraField->value; ?></span> </li> <?php endif; ?> <?php endforeach; ?> </ul> <div class="clr"></div> </div> <?php endif; ?> Now I want to check the name field for a value and if it is Closing Date print out something different from the value in value, so I've changed the red code to <span class="catItemExtraFieldsLabel"><?php if $extraField->name ="Closing date"; ?></span> { <span class="catItemExtraFieldsLabel"><?php echo $extraField->name; ?></span> <span class="catItemExtraFieldsValue"><?php echo $this->item->publish_down, JText::_('K2_DATE_FORMAT_LC2')); ?></span> } else { <span class="catItemExtraFieldsLabel"><?php echo $extraField->name; ?></span> <span class="catItemExtraFieldsValue"><?php echo $extraField->value; ?></span> } <?php endif; ?> Doesn't work though, tried a heap of variations but can't figure it out. Thanks Marc Hey im trying to come up with a simple script that shows the databases in my localhost then create a insert box with a "submit button" and when i type in any name lets just say test1234 it should show up in the list and i want it to show the results below such as you successfully added a database. I have started this not too long ago but i've had experience with previous database adding and i just threw it in there to get me started. Here is my code as of now. By the way I am using PHP my admin and i have it so it creates it to the list but it does it automatically but i want it to add it by me typing it in the text box and when i hit the submit button it shows up in the list and below as well. Code: [Select] <? $connection = @mysql_connect("localhost", "root", "") or die(mysql_error());; $dbs = @mysql_list_dbs($connection)or die(mysql_error()); $db_list="<ul>"; $i =0; while ($i < mysql_num_rows($dbs)){ $db_names[$i] = mysql_tablename($dbs,$i); $db_list .="<li>$db_names[$i]"; $i++; } $db_list .="</ul>"; ?> <HTML> <HEAD> <TITLE>MySQL Databases</TITLE> </HEAD> <P><strong>Databases on localhost</strong>:</p> <? echo "$db_list"; ?> </BODY> </HTML> <HTML> <HEAD> <TITLE>Adding a Database to MySQL</TITLE> </HEAD> <BODY><H3> <FORM METHOD="post" ACTION="pretask.php"> <P>Database Name <INPUT TYPE="text" NAME="val1" SIZE=10></P> <P><INPUT TYPE="submit" NAME="submit" VALUE="Add database"></P> </FORM> </BODY></H3> </HTML> </font> <? $connection = @mysql_connect("localhost", "root", "") or die(mysql_error()); if ($connection) { $msg = "YES!"; } $sql = "CREATE DATABASE my_music "; $result = @mysql_query($sql,$connection) or die(mysql_error()) ?> <HTML> <BODY> <h1><center><? echo "$msg"; ?></h1></center> </BODY> </HTML> I am building a system where people are able to choose a food menu, they can modify each dish if needed and then say how many of each item they need for the order. So when they pick the breakfast menu, it selects all starters, mains and desserts. Should something need to be edited (sausages changed to vegitarian sausages) the user should be able to overtype what has been populated from the select statement) When the form is submitted, the order needs to be linked to the current job which is in a field ssm_job.job_id but the changes to the menu items need to be stored somewhere for the current order only. This obviously needs to be accessile later so further changes can be made if needed. I am not sure of the best way to go about this as i am trying not to save every item in another database (just trying to be tidy) Any advise on this would be greatly appreciated. I have had help on this project in relation to an sql and the background can be found below, the help has been amazing but as this is a very different question i thought i would start a new thread.
Kind Regards Hello. I want to check if an e-mail address is stored into a database. I've found this type of code: Code: [Select] function verifmail($email) { db_connect(); if (mysql_num_rows(mysql_query("SELECT email FROM users WHERE email='$email';"))) { return 1; } else { return 0; } } Is it correct ? Is there any other way? I have a set of 26 checkboxes that are stored in an array. The below code dumps the comma separated array values into a single record column. Code: [Select] // dumping the type of job checkboxes $type = $_POST['type']; var_dump($type); // Setting up a blank variable to be used in the coming loop. $allStyles = ""; // For every checkbox value sent to the form. foreach ($type as $style) { // Append the string with the current array element, and then add a comma and a space at the end. $allTypes .= $style . ", "; } // Delete the last two characters from the string. $allTypes = substr($allTypes, 0, -2); // end dumping the type of job checkboxes This works great! But my needs go past this. The record can be edited. So I need to get the values back out of the database and populate the correct check boxes for editing. The checkboxes are generated by this code. Code: [Select] <div class="text">Job Type (<a class="editlist" href="javascript:editlist('listedit.php?edit=typelist',700,400);">edit list</a>):</div> <div class="field"> <?PHP $connection=mysql_connect ("localhost", "name", "pass") or die ("I cannot connect to the database."); $db=mysql_select_db ("database", $connection) or die (mysql_error()); $query = "SELECT type FROM typelist ORDER BY type ASC"; $sql_result = mysql_query($query, $connection) or die (mysql_error()); $i=1; echo '<table valign="top"><tr><td>'; while ($row = mysql_fetch_array($sql_result)) { $type = $row["type"]; if ($i > 1 && $i % 26 == 0) echo '</td><td>'; else if ($i) echo ''; ++$i; echo "<input style='font-size:10px;' name='type[]' type='checkbox' value='$type'><span style='color:#000;'>$type</span></input><br/>"; } echo '</td></tr></table>'; ?> </div> I am having a problem getting these checkboxes repopulated. I think that a way to do this is to: Get the values out of the database in a array Compare the values the list of all possible values. Check the boxes of the values confirmed to be there. I didn't build the application so I need help to work through this problem. I'm creating a game where 2 people, on separate computers can battle each other. I'm doing it turn-based, like Final Fantasy but I came across a problem, how do I check if the other person has ended their turn? I was thinking that I would have an area in my database that would be true/false if it was their turn. So if the first person ends their turn, their 'turn' becomes false in the database, but how, on the other users computer, do I constantly check if the 'turn' has become false so that they may now play without having to keep updating the browser. Or is their a better way to do something like this? I'm trying to get a nickname system working for a little project, but I've run into a problem. This code that I wrote here gets the users nickname from their username: Code: (php) [Select] $user = $_GET['user']; @mysql_select_db("db") or die( "Unable to select database"); $query = mysql_query("SELECT `nickname` FROM `info` WHERE `username`='$user'"); WHILE($rows = mysql_fetch_array($query)): $nickname = $rows['nickname']; echo $nickname; endwhile; ?> But the problem is, if the user doesn't have a nickname then it just turns up blank. How can I check if the output of $nickname is blank and if it is then just return their normal name ($user) Thanks Oh, and how to you format php code to look like php code on these forums? Hi, how to make this form + php to make it work this way from the drop-down box choose the database title, or ID, and then the form in which I write some content will be added to the database that matches the title or ID that is sometimes so as not to invade a mistake and content that I type is not added to another title or ID I created something like this: <?php $db = new mysqli('localhost','xxxxx','xxxx','xxxx'); mysqli_query($db,'SET NAMES `utf8`'); $sqlnowe = mysqli_query($db,'SELECT * FROM `cc` ORDER BY `id` DESC '); ?> <form action="formu.php" method="post"> <select name="tytul"> <?php while ($rownowe = mysqli_fetch_array($sqlnowe)) { ?> <option><? echo $rownowe['tytul']; ?></option><?php };?> </select> opismax:<br /> <input type="text" name="opismax" /><br /> <textarea name="opismax" cols="50" rows="10">Proszę, wpisz tutaj jakiś komentarz...</textarea> <input type="submit" value="dodaj" /> </form> <?php // odbieramy dane z formularza $tytul = $_POST['tytul']; $opismax = $_POST['opismax']; if($tytul and $opismax) { // dodajemy rekord do bazy $ins = mysqli_query("INSERT tytul='$tytul' INTO publications SET opismax='$opismax'"); if($ins) echo "Good"; else echo "Bad"; } ?> Hey anybody can please guide me where I m wrong in this php code please reply here is the code Only variables should be passed by reference on line 3 in that function area function get_file_extension($file_name) main error I guess is the explode one Aand yeah I checked adding $file_name = $_FILES['fld']; before the function is started <?php require("include/dbconn.php"); //$username = mysql_real_escape_string($_POST['username1']); function get_file_extension($file_name) { return end(explode('.',$file_name)); } function errors($error){ if (!empty($error)) { $i = 0; while ($i < count($error)){ $showError.= '<div class="msg-error">'.$error[$i].'</div>'; $i ++;} return $showError; }// close if empty errors } // close function if (isset($_POST['submit'])){ $username = "Sunny"; $date1 = date("d m y"); mysql_select_db("deals") or die(mysql_error()); if($username == "") { //header("location:newnewuser1.php"); echo "Username empty"; //header("location:newnewuser1.php"); } else { function createRandomPassword() { $chars = "abcdefghijkmnopqrstuvwxyz023456789"; srand((double)microtime()*1000000); $i = 0; $pass = '' ; while ($i <= 7) { $num = rand() % 33; $tmp = substr($chars, $num, 1); $pass = $pass . $tmp; $i++; } return $pass; } // Usage $password = createRandomPassword(); //$password = mysql_real_escape_string($_POST['Password']); $date = strtotime("+2hours"); $date1 = date("d-m-y",$date); if(get_file_extension($_FILES["fld"]["name"])!= 'csv') { $error[] = 'Only CSV files accepted!'; echo "Wrong Input"; }//get_file_extension if (!$error) { $tot = 0; $handle = fopen($_FILES["fld"]["tmp_name"], "r"); while (($data = fgetcsv($handle, 1000, ",")) !== FALSE) { for ($c=0; $c < 1; $c++) { try { mysql_query("INSERT INTO newusers (email,password,location,company,name,contact,sourcename,date)VALUES('".mysql_real_escape_string($data[0])."','$password','".mysql_real_escape_string($data[1])."', '".mysql_real_escape_string($data[2])."','".mysql_real_escape_string($data[3])."','".mysql_real_escape_string($data[4])."','$username1','$date1')"); $tot++; } catch(Exception $e) { PRINT 'ERROR:' +$e; } }//for }//while fclose($handle); }// end no error }//close if isset upfile } ?> and this is the form code <form name="f1" action="" method="POST" enctype="multipart/form-data"> <table><tr><td><input type="file" value="Browse" name="fld" onselect="CheckExtension(fld)"><br/></td><td> </td></tr><tr><td>Username :<input type="text" name="username1" value="" id="user" onfocus="validateForm1()"></td> <td><input type="submit" name="submit" value="submit" id="submit"></td></tr></table> </form> Please help me I m stuck in this quite badly Please [attachment deleted by admin] Code: [Select] $product_id=$_GET['product']; $sql500="SELECT * FROM $tbl_name3 WHERE product_id='$product_id'"; $result500=mysql_query($sql500); $num_rows500=mysql_num_rows($result500); while($row500=mysql_fetch_array($result500)){ extract($row500); $review_product_rating_total=$review_product_rating; //What do I do here? } $average_rating=$review_product_rating_total/$num_rows500; I need a way to take a column of product ratings (0-5 in 1/2 increments), add them together, and divide by the number of rows taken from the database. I know how to get the number of rows with mysql_num_rows. But how would I add together the data from database? Hey i have set up a database and im trying to ad records to it also upload an image, when i submit my form it clears the fields but does not add to the database. There are no errors that get outputted or anything so it gives no help I have read thru my code over and over and just cant see why its not adding. Any help would be great <html> <script language="JavaScript"> function validated(){ var matric = document.s.matric.value; var name = document.s.name.value; var course = document.s.course.value; var sem = document.s.sem.value; var semint = parseInt(sem); var tel = document.s.tel.value; var telint = parseInt(tel); var address = document.s.address.value; var picture = document.s.picture.value; if(matric==""){ window.alert("Please enter matric number!"); document.s.matric.focus(); return false; } if(name==""){ window.alert("Please enter student name!"); document.s.name.focus(); return false; } if(course==""){ window.alert("Please enter student course!"); document.s.course.focus(); return false; } if(isNaN(semint)){ window.alert("Please enter student semester!"); document.s.sem.focus(); return false; } if(isNaN(telint)){ window.alert("Please enter contact number!"); document.s.tel.focus(); return false; } if(address==""){ window.alert("Please enter student address!"); document.s.address.focus(); return false; } if(picture==""){ window.alert("Please enter student picture!"); document.s.picture.focus(); return false; } } </script> <body> <table width="100%" border="0" cellspacing="0" cellpadding="2"> <tr> <td align="center"><h1><font color="#0000FF" face="Arial">ADD STUDENT PROFILE</font></h1></td> </tr> </table> <br><br> <table width="100%" border="0" cellspacing="0" cellpadding="2"> <tr> <td align="right"><font size="1" face="Arial"><a href='main.php'>Student List</a></font></td> </tr> </table> <br><br> <form method="post" action="student_form.php" enctype="multipart/form-data" name='s' onsubmit='return validated()';> <table width="70%" border="0" align="center" cellpadding="2" cellspacing="2"> <tr> <td width="33%" align="right"><font size="2" face="Arial"><strong>Matric Number</strong></font></td> <td width="5%" align="center">:</td> <td width="62%"><input type='text' name='matric' size=30 maxlength=15></td> </tr> <tr> <td align="right"><font size="2" face="Arial"><strong>Name</strong></font></td> <td align="center">:</td> <td><input type='text' name='name' size=30 maxlength=50></td> </tr> <tr> <td align="right"><font size="2" face="Arial"><strong>Course</strong></font></td> <td align="center">:</td> <td><input type='text' name='course' size=30 maxlength=50></td> </tr> <tr> <td align="right"><font size="2" face="Arial"><strong>Semester</strong></font></td> <td align="center">:</td> <td><input type='text' name='sem' size=5 maxlength=2></td> </tr> <tr> <td align="right"><font size="2" face="Arial"><strong>Sex</strong></font></td> <td align="center">:</td> <td><input type='radio' name='sex' value='Male' checked> Male <input type='radio' name='sex' value='Female'> Female</td> </tr> <tr> <td align="right"><font size="2" face="Arial"><strong>Contact Number</strong></font></td> <td align="center">:</td> <td><input type='text' name='tel' size=30 maxlength=50></td> </tr> <tr> <td align="right"><font size="2" face="Arial"><strong>Address</strong></font></td> <td align="center">:</td> <td><input type='text' name='address' size=50 maxlength=100></td> </tr> <tr> <td align="right"><font size="2" face="Arial"><strong>Picture</strong></font></td> <td align="center">:</td> <td> <input type="hidden" name="MAX_FILE_SIZE" value="10485760"> <input type="file" name="picture" size="40"> <font size="1" face="Arial"> Maxsize 1MB</font> </td> </tr> <tr> <td> </td> <td> </td> <td><input type='submit' name="submit" value='Submit'> <input type='reset' value='Reset'></td> </tr> </table> <?php if ($submit) { include 'db_connect.php'; $data = addslashes(fread(fopen($picture, "r"), filesize($picture))); $pjpeg="image/pjpeg"; $jpeg="image/jpeg"; $gif="image/gif"; $png="image/png"; $bmp="image/bmp"; if ($picture_type == $pjpeg OR $picture_type == $jpeg OR $picture_type == $gif OR $picture_type == $png OR $picture_type == $bmp) { $sql="INSERT INTO info (matric, name, course, sem, sex, tel, address ,bin_data,filename,filesize,filetype) VALUES ('$matric', '$name', '$course', '$sem', '$sex', '$tel', '$address', '$data','$picture_name','$picture_size','$picture_type')"; $result=mysql_query($sql); header ("Location: main.php"); } else{ echo "<script language='JavaScript'>"; echo "window.alert('Error! You only can upload jpeg, gif, png, bmp file type.')"; echo "</script>"; } } ?> </form> </body> </html> Thanks, I am stuck on inserting the data into mysql via php. Its a foreach method and I am just stuck Code: [Select] include '../Database/take_an_exam.php'; $intNum = 1; $intnumber = 1; while( $info = mysql_fetch_array( $sqll )){ echo "<input type='hidden' name=\"Que_ID\" value=\"{$info['Que_ID']}\" /> "; echo " $intNum, {$info['Que_Question']} <br />\n"; $intNum++; for ($i =1; $i < 5; $i++) { echo "<input type=\"checkbox\" name=\"choice[{$info['Que_ID']}.$i][]\" />{$info['Que_Choice'.$i]}<br />\n"; $intnumber++; } } ?> <input type="submit" value="submit" name="submit"> This is my form that can output the Question and its choices. I am echoing this on to the next page, the code is shown below Code: [Select] <?PHP session_start(); $name= $_SESSION['username1']; echo "User ID: $name <br/>"; $gender = $_POST["choice"]; $que_ID = $_POST["Que_ID"]; foreach ($gender as $key => $array) { if($array) { $val=1; } ///echo "Question ID and the choice ID:$key. Value is:.$val <br />\n"; $well = array(floor($key), substr(strstr($key, '.'), 1)); //$well = array(floor($key), $key - floor($key)); echo "Question ID: $well[0], "; echo "Choice: $well[1]. value: $val<br/>"; } //}break; ?> Now, instead of echoing. I just want to add this to mysql. The layout of my sql is Que_ID, Ans_Choice1, Ans_Choice2, Ans_Choice3, Ans_Choice4, Use_ID When I echo my php my outcome is Quote User ID: 2 Question ID= 1 Choice 1 Question ID= 2 Choice 3 Question ID= 3 Choice 4 Question ID= 4 Choice 1 I embed YouTube videos on my site.
The problem is, the videos often get deleted by the user, which means that the videos obviously become unavailable on my site.
This happens often, which means that I regularly have to look through my site, find unavailable videos and delete them.
I am looking for an automated way of doing this.
Here's what I've got so far:
<?php
$con = mysql_connect("", "", "");
mysql_select_db("");
$qry = "SELECT video_id FROM `table`";
$run = mysql_query($qry, $con);
while($row = mysql_fetch_assoc($run)) {
$headers = get_headers("http://gdata.youtube.com/feeds/api/videos/{$row['video_id']}");
if(!strpos($headers[0], '200')) {
echo "Unavailable: {$row['video_id']}<br />";
return false;
}
else {
echo "Available: {$row['video_id']}<br />";
}
}
?>
The code's very slow and doesn't work. It produces a list of available videos and then stops when it finds one that is unavailable. The problem is, the video that it says is unavailable is usually available, which means that it's getting it wrong.
Is there a faster, more accurate way of doing it?
Basically, I want a fast and accurate way of checking the availability of the thousands of videos in my database and I want to produce a list of the ones that are unavailable.
Like many projects, this started small, and keeps growing! I am building a photo gallery and would now like to add labels/captions below each photo, HOWEVER, I really don't want to have to build a database and do all the related coding. (I know how, but its a PITA.) To display photos in the gallery, I read in files from my "images/" directory, store them in a simple array, and then iterate through the array to display the thumbnails in a gallery. I am wondering if there would be an easy way that I could merge photo metadata (e.g. a brief caption) with my array or something like that? If I could type up the file names and a brief description/caption in a Text File or Spreadsheet and then somehow slurp that into my code and merge it with my array or something like that, then I would be willing to type up captions. (This is for like 600 photos which is why I don't want to do a database as data entry in phpMyAdmin is a PITA!) Any suggestions how I could do this and add a "cherry on the top" of this mini project I am doing for my co-workers? Thanks!
hello, i am hoping someone can help, i have a form that has a body and title fields and then sends to this function below. it all works fine, but when i add a image or a link it stores it in the text field of the db like this: <IMG alt=\"\" src=\"/public/images/231781538234094.jpg\" width=796></P> this is what it should be /public/images/231781538234094.jpg so when i view the image it doesent show it and i right click the image and goto image properties and i get this: http://test.cyberglide.co.uk/%22public/images/231781538234094.jpg/%22 Code: [Select] function content_update() { $title = mysql_real_escape_string($_POST['title']); $body = mysql_real_escape_string($_POST['body']); $page = mysql_real_escape_string($_POST['page']); $location = mysql_real_escape_string($_POST['location']); $id = mysql_real_escape_string($_POST['id']); $sql = "UPDATE content SET title = '$title', body = '$body', page = '$page', location = '$location' WHERE id = '$id'"; $res = mysql_query($sql) or die(mysql_error()); echo "<script>window.location='content.php'</script>"; } if i manually edit it on the db it works fine please help, many thanks. Good day guyz... I dont know why this code does not work... the connection is fine, the database is also fine it works with the localhost, when i upload it on free hosting site like 000webhost and infinityfree this is not working.. the database on free hosting site are also working also the connection is code, just only this code i cant figure it out.. else{
$sql = "INSERT INTO .$name (TrxID, ProductName, Price, ProductImage, PurchasedDate, PurchasedStatus) VALUES (?, ?, ?, ?, ?, ?)";
$stmt = mysqli_stmt_init($conn2);
if(!mysqli_stmt_prepare($stmt, $sql)){
echo 'connection error';
}
mysqli_stmt_bind_param($stmt, "isssss", $val, $PName, $Price, $PImage, $PDate, $PStatus);
mysqli_stmt_execute($stmt);
mysqli_stmt_close($stmt);
header("location: ../customer/cart.html");
exit();
}
}
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