PHP - Add Images Help (explaination Under Code)
<?php /** * This is the model class for table "student_docs". * * The followings are the available columns in table 'student_docs': * @property integer $student_docs_id * @property string $doc_category_id * @property string $title * @property string $student_docs_desc * @property string $student_docs_path */ class StudentDocs extends CActiveRecord { /** * Returns the static model of the specified AR class. * @param string $className active record class name. * @return StudentDocs the static model class */ public static function model($className=__CLASS__) { return parent::model($className); } /** * @return string the associated database table name */ public function tableName() { return 'student_docs'; } /** * @return array validation rules for model attributes. */ public function rules() { // NOTE: you should only define rules for those attributes that // will receive user inputs. return array( array('student_docs_path, doc_category_id, title,student_docs_submit_date', 'required','message'=>""), array('student_docs_desc', 'length', 'max'=>50), array('student_docs_path', 'file', 'types'=>'jpeg, jpg, pdf, txt, doc,docx, gif, png', 'maxSize'=>1024*1024*2, 'tooLarge'=>'The document was larger than 2MB. Please upload a smaller document.',), //array('title','CRegularExpressionValidator','pattern'=>'/^[a-zA-Z& ]+([-]*[a-zA-Z0-9 ]+)*$/','message'=>''), // The following rule is used by search(). // Please remove those attributes that should not be searched. array('student_docs_id, student_docs_desc, student_docs_path, doc_category_id,student_docs_submit_date, title', 'safe', 'on'=>'search'), ); } /** * @return array relational rules. */ public function relations() { // NOTE: you may need to adjust the relation name and the related // class name for the relations automatically generated below. return array( ); } /** * @return array customized attribute labels (name=>label) */ public function attributeLabels() { return array( 'student_docs_id' => 'Student Docs', 'student_docs_desc' => 'Document Description', 'student_docs_path' => 'Document', 'doc_category_id' => 'Document Category', 'title' => 'Title', 'student_docs_submit_date'=>'Submit Date', ); } /** * Retrieves a list of models based on the current search/filter conditions. * @return CActiveDataProvider the data provider that can return the models based on the search/filter conditions. */ public function search() { // Warning: Please modify the following code to remove attributes that // should not be searched. $criteria=new CDbCriteria; $criteria->compare('student_docs_id',$this->student_docs_id); $criteria->compare('doc_category_id',$this->doc_category_id,true); $criteria->compare('title',$this->title,true); $criteria->compare('student_docs_desc',$this->student_docs_desc,true); $criteria->compare('student_docs_path',$this->student_docs_path,true); return new CActiveDataProvider($this, array( 'criteria'=>$criteria, )); } }Hello everyone! Please help save me because I am about to loose my mind. First- I have a database for a company I have been working on and for this database one of the things they would like is to have a tab to add documents to a specific clients database profile. I have done all of that and added the tab to add documents however, it will not allow me to add the document. I have attached a picture to show you what the error is. Basically everything else is working fine- I give the file a name, description, date of upload, and all of that gives me the green arrow beside it saying its fine however, when I add the file (any file extension) it will give me a red "x" beside the file path saying it can't be added. (see picture adddocument.png to see the display). Here is also the code that is controlling the document upload page- maybe I am missing something. Maybe you will be able to catch it before I will- I have been staring at this for days now trying to get it to work! Thanks for the help in advance! Attached Files Add Document.png 15.22KB 0 downloads Similar Tutorials$page = (isset($_GET["page"]))?$_GET["page"] : 1 ; This is from a pagination script, and I am wondering what the ? means here, and what the : does. Hey, I'm finally getting around to teaching myself OO PHP, I been working on a database connetcion class and based it on a tutorial I found on the net. However, it doesn't explain it too well can anyone help me please. I know what is going on right up until " $this->_database_connection" and "$this->_database_connection_select", at first I had the coding as "$this->databaseConnection = mysql_pconnect($this->databaseHostname, $this->databaseUsername, $this->databasePassword) or trigger_error(mysql_error(),E_USER_ERROR);" because I thought it was saysing the this function database connection should connect to the mysql database using these variables. But it didn't seem to working, got it fixed now using after looking at the tutorial but there is no explaination to why this works and mine didn't. I understand that databaseHostname for example is being defined by the object, on the third bit of coding. Actually thinking about it, have I also defined in the first bit of coding. I haven't included the SQL statement that retrives data, just so you know. If anyone can help I will greatly appericate it Code: [Select] <?php # CONNECTION DETAILS THAT ARE THEN PASSED THROUGH TO "dbConnecxt" $databaseHostname = "localhost"; $databaseUsername = "username"; $databasePassword = "password"; $databaseName = "databasename"; ?> Code: [Select] <?php require_once ("config/config.php"); class databaseConnectionClass { public $databaseHostname; public $databaseUsername; public $databasePassword; public $databaseName; # MAIN CONNECTION TO THE DATABASE, PASSING THE public function databaseConnection($objDatabaseConnect) { $this->_database_connection = mysql_pconnect($this->databaseHostname, $this->databaseUsername, $this->databasePassword) or trigger_error(mysql_error(),E_USER_ERROR); return $this->_database_connection; } # SELECTS THE DATABASE WE WANT public function databaseConnectionSelect() { $this->_database_connection_select = mysql_select_db($this->databaseName, $this->_database_connection); return $this->_database_connection_select; } # CALL ALL THE DATABASE CONNECTION OBJECTS public function databaseConnectionProcess($objDatabaseConnect) { $objDatabaseConnect->databaseConnection($objDatabaseConnect); $objDatabaseConnect->databaseConnectionSelect($objDatabaseConnect); } # BUILDS A OBJECT METHOD public function databaseConnectionMain($objDatabaseConnect) { $objDatabaseConnect->databaseConnectionProcess($objDatabaseConnect); } } ?> Code: [Select] <?php #THIS CLASS CREATES AND OBJECT, WHICHS SETS THE OBJECT TO EQUAL THE INFO FROM "dbConnectClass.php" require_once ("dbConnectClass.php"); $objDatabaseConnect = new databaseConnectionClass(); $objDatabaseConnect->databaseHostname = $databaseHostname; $objDatabaseConnect->databaseUsername = $databaseUsername; $objDatabaseConnect->databasePassword = $databasePassword; $objDatabaseConnect->databaseName = $databaseName; $objDatabaseConnect->databaseConnectionMain($objDatabaseConnect); ?> I have a phppage which needs 1) to load images ,image paths and image name from a different server,when the page is loaded 2)Details automatically saved to the database. can we do this with php(i mean can we take datas from a different server through php code) Hello everyone.. This is the first PHP script I've written and was hoping to get some feedback on any possible issues with it. I've pieced this together in an attempt to download remote images and store them on my server, instead of hotlinking images. The code will be used for a forum, called up by a BBCode tag. (The user will place an image URL into the BBCode, which will transfer to this PHP script). Again, this is the first time I've coded anything in PHP and was hoping to get some pointers on anything that needs changing.. thanks <?php $url = $_GET['url']; $url_path = parse_url($url, PHP_URL_PATH); $name = basename($url_path); $FileExt= substr($name, -3); $FileTypeMIME= array("jpg" => "image/jpeg", "png" => "image/png", "gif" => "image/gif"); $ContentType= $FileTypeMIME[$FileExt]; if (empty($ContentType)) die("You are not allowed to access this file!"); header("Content-Type: " . $ContentType); $save = "../images/". strtolower($name); function wtf_image ($file) { switch($FileTypeMIME[$FileExt]){ case "image/jpeg": $im = imagecreatefromjpeg($file); //jpeg file imagejpeg($im, $save, 0, NULL); //save jpeg file break; case "image/gif": $im = imagecreatefromgif($file); //gif file imagegif($im, $save, 0, NULL); //save gif file break; case "image/png": $im = imagecreatefrompng($file); //png file imagePNG($im, $save, 0, NULL); //save png file break; } return $im; } if (file_exists($save)) { readfile($save); } else { chmod($save,0755); $image = wtf_image($url); //Runs wtf_image function on $url imagedestroy($image); readfile($save); } ?> I am looking for the code|script to put on my update.php that allows uploaded images to be saved in my database (chmod) 777? PLEASE HELP! I have this script from http://lampload.com/...,view.download/ (I am not using a database) I can upload images fine, I can view files, but I want to delete them. When I press the delete button, nothing happens
http://www.jayg.co.u...oad_gallery.php
<form>
<?php $dir = dirname(__FILENAME__)."/images/gallery" ; $files1 = scandir($dir); foreach($files1 as $file){ if(strlen($file) >=3){ $foil = strstr($file, 'jpg'); // As of PHP 5.3.0 $foil = $file; $pos = strpos($file, 'css'); if ($foil==true){ echo '<input type="checkbox" name="filenames[]" value="'.$foil.'" />'; echo "<img width='130' height='38' src='images/gallery/$file' /><br/>"; // for live host //echo "<img width='130' height='38' src='/ABOOK/SORTING/gallery-dynamic/images/gallery/ $file' /><br/>"; } } }?> <input type="submit" name="mysubmit2" value="Delete"> </form>
any ideas please?
thanks
I have the following code in html: <html> <head> <script type="text/javascript"> <!-- function delayer(){ window.location = "http://VARIABLEVALUE.mysite.com" } //--> </script> <title>Redirecting ...</title> </head> <body onLoad="setTimeout('delayer()', 1000)"> <script type="text/javascript"> var sc_project=71304545; var sc_invisible=1; var sc_security="9c433fretre"; </script> <script type="text/javascript" src="http://www.statcounter.com/counter/counter.js"></script><noscript> <div class="statcounter"><a title="vBulletin statistics" href="http://statcounter.com/vbulletin/" target="_blank"><img class="statcounter" src="http://c.statcounter.com/71304545/0/9c433fretre/1/" alt="vBulletin statistics" ></a></div></noscript> </body> </html> Is a basic html webpage with a timer redirect script and a stascounter code. I know a bit about html and javascript, but almost nothing about php. My question is: How a can convert this html code into a php file, in order to send a variable value using GET Method and display this variable value inside the javascript code where says VARIABLEVALUE. Thanks in adavance for your help. Hi, I have some code which displays my blog post in a foreach loop, and I want to add some social sharing code(FB like button, share on Twitter etc.), but the problem is the way I have my code now, creates 3 instances of the sharing buttons, but if you like one post, all three are liked and any thing you do affects all of the blog post. How can I fix this? <?php include ("includes/includes.php"); $blogPosts = GetBlogPosts(); foreach ($blogPosts as $post) { echo "<div class='post'>"; echo "<h2>" . $post->title . "</h2>"; echo "<p class='postnote'>" . $post->post . "</p"; echo "<span class='footer'>Posted By: " . $post->author . "</span>"; echo "<span class='footer'>Posted On: " . $post->datePosted . "</span>"; echo "<span class='footer'>Tags: " . $post->tags . "</span>"; echo ' <div class="addthis_toolbox addthis_default_style "> <a class="addthis_button_facebook_like" fb:like:layout="button_count"></a> <a class="addthis_button_tweet"></a> <a class="addthis_counter addthis_pill_style"></a> </div> <script type="text/javascript">var addthis_config = {"data_track_clickback":true};</script> <script type="text/javascript" src="http://s7.addthis.com/js/250/addthis_widget.js#username=webguync"></script>'; echo "</div>"; } ?> hey gurus, i am a newbie php coder.. i am learning by example. what i am trying to do is write a piece of code which will alter 3 tables (user, bonus_credit, bonus_credit_usage) ---------------------------------------------------------------- the table structure that will be used is as follows: user.bonus_credit user.ID bonus_credit.bonusCode bonus_credit.qty bonus_credit.value bonus_credit_usage.bonusCode bonus_credit_usage.usedBy ---------------------------------------------------------------- so lets say, in bonus_credit i have the following bonusCode = 'facebook' (this is the code they have to type to redeem the bonus qty = '10' ( number of times the bonusCode can be redeemed, but same person can't redeem it more than once) value = '5' (this is the amount of bonus_credit for each qty) Now, I need to write a code that check to see if the code has been redeemed in the bonus_credit_usage table and if the user.ID exists in this table as bonus_code_usage.usedBy, then give an error that its already been used and if it hasn't been used, then subtract 1 from qty, add ID to usedBy and then add the value to the bonus_credit ----------------------- i have started the steps just to create a simple textbox and entering a numeric value to bonus_credit, and that works.. but now i have to use JOIN and IF and ELSE.. which is a little too advanced for me.. so i'd appreciate a guide as i write the code. if(isset($_REQUEST['btnBonus'])) { $bonus_credit = addslashes($_REQUEST['bonusCode']); $query = "update user set bonus_credit=bonus_credit+'".$bonus_credit."' where id='".$_SESSION['SESS_USERID']."'"; echo "<script>window.location='myreferrals.php?msgs=2';</script>"; mysql_query($query) or die(mysql_error()); } Advance thank you. Can you help please. The error..... Warning: mysql_fetch_assoc() expects parameter 1 to be resource, string given in C:\wamp\www\test_dabase.php on line 24 code. Code: [Select] <?php //database connection. $DB = mysql_connect("localhost","root") or die(mysql_error()); if($DB){ //database name. $DB_NAME="mysql"; //select database and name. $CON=mysql_select_db($DB_NAME,$DB)or die(mysql_error()."\nPlease change database name"); // if connection. }if($CON){ //show tables. $mysql_show="SHOW TABLES"; //select show and show. $mysql_select2="mysql_query(".$mysql_show.") or die(mysql_error())"; } //if allowed to show. if($mysql_select2){ //while it and while($data=mysql_fetch_assoc($mysql_select2)){ //show it. echo $data; } } ?> Hi, I need to insert some code into my current form code which will check to see if a username exist and if so will display an echo message. If it does not exist will post the form (assuming everything else is filled in correctly). I have tried some code in a few places but it doesn't work correctly as I get the username message exist no matter what. I think I am inserting the code into the wrong area, so need assistance as to how to incorporate the username check code. $sql="select * from Profile where username = '$username'; $result = mysql_query( $sql, $conn ) or die( "ERR: SQL 1" ); if(mysql_num_rows($result)!=0) { process form } else { echo "That username already exist!"; } the current code of the form <?PHP //session_start(); require_once "formvalidator.php"; $show_form=true; if (!isset($_POST['Submit'])) { $human_number1 = rand(1, 12); $human_number2 = rand(1, 38); $human_answer = $human_number1 + $human_number2; $_SESSION['check_answer'] = $human_answer; } if(isset($_POST['Submit'])) { if (!isset($_SESSION['check_answer'])) { echo "<p>Error: Answer session not set</p>"; } if($_POST['math'] != $_SESSION['check_answer']) { echo "<p>You did not pass the human check.</p>"; exit(); } $validator = new FormValidator(); $validator->addValidation("FirstName","req","Please fill in FirstName"); $validator->addValidation("LastName","req","Please fill in LastName"); $validator->addValidation("UserName","req","Please fill in UserName"); $validator->addValidation("Password","req","Please fill in a Password"); $validator->addValidation("Password2","req","Please re-enter your password"); $validator->addValidation("Password2","eqelmnt=Password","Your passwords do not match!"); $validator->addValidation("email","email","The input for Email should be a valid email value"); $validator->addValidation("email","req","Please fill in Email"); $validator->addValidation("Zip","req","Please fill in your Zip Code"); $validator->addValidation("Security","req","Please fill in your Security Question"); $validator->addValidation("Security2","req","Please fill in your Security Answer"); if($validator->ValidateForm()) { $con = mysql_connect("localhost","uname","pw") or die('Could not connect: ' . mysql_error()); mysql_select_db("beatthis_beatthis") or die(mysql_error()); $FirstName=mysql_real_escape_string($_POST['FirstName']); //This value has to be the same as in the HTML form file $LastName=mysql_real_escape_string($_POST['LastName']); //This value has to be the same as in the HTML form file $UserName=mysql_real_escape_string($_POST['UserName']); //This value has to be the same as in the HTML form file $Password= md5($_POST['Password']); //This value has to be the same as in the HTML form file $Password2= md5($_POST['Password2']); //This value has to be the same as in the HTML form file $email=mysql_real_escape_string($_POST['email']); //This value has to be the same as in the HTML form file $Zip=mysql_real_escape_string($_POST['Zip']); //This value has to be the same as in the HTML form file $Birthday=mysql_real_escape_string($_POST['Birthday']); //This value has to be the same as in the HTML form file $Security=mysql_real_escape_string($_POST['Security']); //This value has to be the same as in the HTML form file $Security2=mysql_real_escape_string($_POST['Security2']); //This value has to be the same as in the HTML form file $sql="INSERT INTO Profile (`FirstName`,`LastName`,`Username`,`Password`,`Password2`,`email`,`Zip`,`Birthday`,`Security`,`Security2`) VALUES ('$FirstName','$LastName','$UserName','$Password','$Password2','$email','$Zip','$Birthday','$Security','$Security2')"; //echo $sql; if (!mysql_query($sql,$con)) { die('Error: ' . mysql_error()); } else{ mail('email@gmail.com','A profile has been submitted!',$FirstName.' has submitted their profile',$body); echo "<h3>Your profile information has been submitted successfully.</h3>"; } mysql_close($con); $show_form=false; } else { echo "<h3 class='ErrorTitle'>Validation Errors:</h3>"; $error_hash = $validator->GetErrors(); foreach($error_hash as $inpname => $inp_err) { echo "<p class='errors'>$inpname : $inp_err</p>\n"; } } } if(true == $show_form) { ?> I use this type of a code to send automatic emails from my website: Code: [Select] $headers = ; $headers .= ; $to = ; Click here to go to Google. ", $headers); I am having hard time figuring out how to do hyperlink on words (like here). If I do something like this: Code: [Select] <a href='http://www.google.com'>here</a> it spits out that exact thing out. Thanks you for your input Can you help me integrate this code :
<form method="post" action="submit.php"> <input type="checkbox" class="required" /> Click to check <br /> <input disabled="disabled" type='submit' id="submitBtn" value="Submit"> </form>In to this Contact Form code, please? <form action="../page.php?page=1" method="post" name="contact_us" onSubmit="return capCheck(this);"> <table cellpadding="5" width="100%"> <tr> <td width="10" class="required_field">*</td> <td width="80">Your Name</td> <td><input type="text" name="name" maxlength="40" style="width:400px;/></td> </tr> <tr> <td class="required_field">*</td> <td>Email Address</td> <td><input type="text" name="email" maxlength="40" style="width:400px;/></td> </tr> <tr> <td></td> <td>Comments:</td> <td><textarea name="comments" style="width: 400px; height: 250px;"></textarea></td> </tr> </table> </form Hi, Look at this code below: Code: [Select] <?php function outputModule($moduleID, $moduleName, $sessionData) { if(!count($sessionData)) { return false; } $markTotal = 0; $markGrade = 0; $weightSession = 0; $grade = ""; $sessionsHTML = ""; foreach($sessionData as $session) { $sessionsHTML .= "<p><strong>Session:</strong> {$session['SessionId']} <strong>Session Mark:</strong> {$session['Mark']}</strong> <strong>Session Weight Contribution</strong> {$session['SessionWeight']}%</p>\n"; $markTotal += round($session['Mark'] / 100 * $session['SessionWeight']); $weightSession += ($session['SessionWeight']); $markGrade = round($markTotal / $weightSession * 100); if ($markGrade >= 70) { $grade = "A"; } else if ($markGrade >= 60 && $markGrade <= 69) { $grade = "B"; } else if ($markGrade >= 50 && $markGrade <= 59) { $grade = "C"; } else if ($markGrade >= 40 && $markGrade <= 49) { $grade = "D"; } else if ($markGrade >= 30 && $markGrade <= 39) { $grade = "E"; } else if ($markGrade >= 0 && $markGrade <= 29) { $grade = "F"; } $moduleHTML = "<p><br><strong>Module:</strong> {$moduleID} - {$moduleName} <strong>Module Mark:</strong> {$markTotal} <strong>Mark Percentage:</strong> {$markGrade} <strong>Grade:</strong> {$grade} </p>\n"; return $moduleHTML . $sessionsHTML; } $output = ""; $studentId = false; $courseId = false; $moduleId = false; while ($row = mysql_fetch_array($result)) { if($studentId != $row['StudentUsername']) { //Student has changed $studentId = $row['StudentUsername']; $output .= "<p><strong>Student:</strong> {$row['StudentForename']} {$row['StudentSurname']} ({$row['StudentUsername']})\n"; } if($courseId != $row['CourseId']) { //Course has changed $courseId = $row['CourseId']; $output .= "<br><strong>Course:</strong> {$row['CourseId']} - {$row['CourseName']} <strong>Course Mark</strong> <strong>Grade</strong> <br><strong>Year:</strong> {$row['Year']} </p>\n"; } if($moduleId != $row['ModuleId']) { //Module has changed if(isset($sessionsAry)) //Don't run function for first record { //Get output for last module and sessions $output .= outputModule($moduleId, $moduleName, $sessionsAry); } //Reset sessions data array and Set values for new module $sessionsAry = array(); $moduleId = $row['ModuleId']; $moduleName = $row['ModuleName']; } //Add session data to array for current module $sessionsAry[] = array('SessionId'=>$row['SessionId'], 'Mark'=>$row['Mark'], 'SessionWeight'=>$row['SessionWeight']); } //Get output for last module $output .= outputModule($moduleId, $moduleName, $sessionsAry); //Display the output echo $output; } } } ?> This code allallows me to make calculations and display a student's course and linked with it the course the modules in the course and linked with modules are all the sessions. It is able to display what marks each student have got for each module and session. Now look at code below, it is able to display modules and in those modules the sessions that link to those modules: Code: [Select] <?php if($moduleId != $row['ModuleId']) { //Module has changed if(isset($sessionsAry)) //Don't run function for first record { //Get output for last module and sessions $output .= outputModule($moduleId, $moduleName, $sessionsAry); } //Reset sessions data array and Set values for new module $sessionsAry = array(); $moduleId = $row['ModuleId']; $moduleName = $row['ModuleName']; } //Add session data to array for current module $sessionsAry[] = array('SessionId'=>$row['SessionId'], 'Mark'=>$row['Mark'], 'SessionWeight'=>$row['SessionWeight']); } What I want to know is how can I do something similar for course so that it picks out the right modules depending on the course it displays. There maybe some code that needs to be added in the function. Can I combine also HTML code in PHP function? For example, can a PHP function include HTML form and the PHP code to handle this form? If yes, this will make my main code much more smaller and readable. If not, is there a way to define an "external macro" like, which allow me to replace pre-defined lines of code with short alias? Alright so I'm attempting to save config data via php. Bellow is the code I currently have, however I'm afraid that when I "flip the switch" and use it that it will error out because of the <?php and ?> tags inside of it... Ideas, suggestions? $config = '../includes/config.php'; $fh = fopen($config, 'w'); $data = ' <?php $dbhost = "'.$database_host.'"; $dbuser = "'.$database_username.'"; $dbpass = "'.$database_password.'"; $dbname = "'.$database_name.'"; $key = "'.$site_key.'"; $cron_key = "'.$database_cron_key.'"; ?> '; fwrite($fh, $data); fclose($fh); How can I make sure that when I submit an new form and new ID (record) is created it is always 4-Digits. record 14 = 0014, record 225 = 0225. Thanks Hello Everyone I am new to php and indeed Web Development. After testing and Playing a bit, I can get the following code to work as two files, the form calling the *.php file to insert into the database, however, I am trying to create one html/php file that displays the form and then executes the php code to insert into the database once user clickes the button. Please can you assist me with the code? I have something horribly wrong and I cannot find it. Code: [Select] <?php> <html> <head> <title>Personal Details</title> </head> <body> <form method="post" action="contactdetails.html"><font face="Arial"> Call Sign:<br> <input name="callsign" size="5" type="text"><br> Surame:<br> <input name="surname" size="30" type="text"><br> First Name:<br> <input name="firstnames" size="30" type="text"><br> Known as:<br> <input name="knownas" size="30" type="text"><br> RSA ID No.:<br> <input name="rsaid" size="13" type="text"><br> Birth Date:<br> <input name="birthdate" size="12" type="text"><br> <input name="Insert" value="Next" type="submit"></form> </font><br> </body> </html> //php to insert data into table $callsign = $_POST['callsign']; $surname = $_POST['surname']; $firstnames = $_POST['firstnames']; $knownas = $_POST['knownas']; $rsaid = $_POST['rsaid']; $birthdate = $_POST['birthdate']; mysql_connect ("localhost", "jredpixm_testuse", "PHPDevelopment") or die ('I cannot connect to the database because: ' .mysql_error()); mysql_select_db ("jredpixm_test"); $query="INSERT INTO personal_details (callsign, surname, firstnames, knownas, rsaid, birthdate)Values ('$callsign', '$surname', '$firstnames', '$knownas', '$rsaid', '$birthdate')"; mysql_query($query) or die ('Error updating Database'); echo "<p>Thanks, your information has been added to the database.</p>"; ?> Regards Allen Hi, this is my first time posting here. I am just delving into PHP and I am learning about foreach loops. I have written code in Notepad++ EXACTLY the way I saw it in a tutorial video I watched (I wish I could show the tutorial video to you, but it is on Lynda.com and you have to pay to watch) I attached the file with my code. The example 1 code works just fine. The example 2 code is the one that is not working for some reason. However, it worked for the guy that wrote it in the video, so I am not sure where I am going wrong? *The comments in green are mainly for myself, I explain things to myself so that I don't forget what the code does forloops.php 1.74KB 2 downloads I would appreciate some help. Thank you!!! Michael Feathers coined the term Legacy Code as being code without automated tests.
Still however Legacy Code evokes a vision in me that it is code that is ugly, old, runs on mainframes, and is probably 3000 lines long, uses globals and questionable code practices.
But say we take this ugly nasty code, and put it very nicely under test, but without doing any refactoring, other than that necessary to be able to put it under test in the first place.
Now that code is under test. But it it still ugly. How would you call ugly code under test?
Would you make a differentiation between old & ugly and modern & pretty code if both are under test?
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