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PHP - Drop Down Data All On The Same Line
When I look at my page in source view the <br> are in the drop down but not working. Here is my php code along with the view source.
This is the main page
<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">
<html xmlns="http://www.w3.org/1999/xhtml">
<head>
<meta http-equiv="Content-Type" content="text/html; charset=iso-8859-1" />
<title>Water Analysis Data(WAD)</title>
<style type="text/css"></style>
</head>
<body>
<p align="center"><a href="../test/index.html">Home</a> | <a href="../test/register.php">Register</a> | <a href="../test/login.php">Login</a> | <a href="../test/tank.php">Add Tank</a> | <a href="../test/fish.php">Add Fish</a> | <a href="../test/water_test.php">Add Water Test</a></p>
<p align="center"> </p>
<table width="810" border="2" align="center">
<tr>
<td><table width="800" border="0" align="center" cellpadding="0" cellspacing="0">
<tr>
<td align="center" bgcolor="#FFFFFF" scope="col"><h2><b>Water Analysis Data(WAD)</b></h2></td>
</tr>
<tr>
<td bgcolor="#FFFFFF">
<form id="form1" method="POST" action="/include/waterparameters/water-parameters.inc.php">
<table border="2" align="center" cellpadding="0">
<tr>
<td><div align="left"><b>Tank Name: </b> </div></td>
<td><div align="left">
<select id="tankname" type='text' name="tankname">
<option>
<?php
include_once '/include/dropdowns/water-parameters-dd.inc.php';
?>
</option>
</select>
</div></td>
</tr>
<tr>
<td><div align="left"><b>Test Date: </b> </div></td>
<td><div align="left">
<input id="date" type="text" name="date" size=25>
</div></td>
</tr>
<tr>
<td><div align="left"><b>Temperatu </b> </div></td>
<td><div align="left">
<input id="temperature" type="text" name="temperature" size=25>
</div></td>
</tr>
<tr>
<td><div align="left"><b>pH: </b> </div></td>
<td><div align="left">
<input id="ph" type="text" name="ph" size=25>
</div></td>
</tr>
<tr>
<td><div align="left"><b>Ammonia: </b> </div></td>
<td><div align="left">
<input id="ammonia" type="text" name="ammonia" size=25>
</div></td>
</tr>
<tr>
<td><div align="left"><b>Nitrite: </b> </div></td>
<td><div align="left">
<input id="nitrite" type="text" name="nitrite" size=25>
</div></td>
</tr>
<tr>
<td><div align="left"><b>Nitrate: </b> </div></td>
<td><div align="left">
<input id="nitrate" type="text" name="nitrate" size=25>
</div></td>
</tr>
<tr>
<td><div align="left"><b>phosphate: </b> </div></td>
<td><div align="left">
<input id="phosphate" type="text" name="phosphate" size=25>
</div></td>
</tr>
<tr>
<td><div align="left"><b>GH: </b> </div></td>
<td><div align="left">
<input id="gh" type="text" name="gh" size=25>
</div></td>
</tr>
<tr>
<td><div align="left"><b>KH: </b> </div></td>
<td><div align="left">
<input id="kh" type="text" name="kh" size=25>
</div></td>
</tr>
<tr>
<td><div align="left"><b>Iron: </b> </div></td>
<td><div align="left">
<input id="iron" type="text" name="iron" size=25>
</div></td>
</tr>
<tr>
<td><div align="left"><b>Potassium: </b> </div></td>
<td><div align="left">
<input id="potassium" type="text" name="potassium" size=25>
</div></td>
</tr>
<tr>
<td><div align="left"><b>Notes: </b> </div></td>
<td><div align="left">
<p>
<textarea id="notes" name="notes" cols="50" rows="10"></textarea>
</p>
</div></td>
</tr>
<tr>
<th colspan=2><p>
<input type="submit" value="Submit"id="submit1" class="submit" name="submit" />
</p></th>
</tr>
</table>
</form></td>
</tr>
<tr>
<td align="center" valign="top" bgcolor="#FFFFFF"><div align="center"><font size=2> © 2014 <a href="http://www.pctechtime.com">PC TECH TIME</a> </font> </div></td>
</tr>
</table></td>
</tr>
</table>
</body>
</html>
This is the include for the page (the echo line for <br> shows in the source but doesn't produce the break
<?php
include_once "/include/db/db.inc.php";
$result = mysqli_query($con,"SELECT tankname FROM tank");
while($row = mysqli_fetch_array($result)) {
echo $row['tankname'];
echo "<br>";
}
mysqli_close($con);
?>
Here is the source when you bring the file up in a browser
<select id="tankname" type='text' name="tankname">
<option>
Tank 1<br>Tank 2<br>Tank 3<br>
</option>
</select>
Similar TutorialsHi, I'm having a bit of trouble with returning data from the database. It brings back the images etc... correctly, but I need to be able to limit it to two images per, line and then to carriage return. Or using CSS to position them with the correct padding etc... I'm very tired and can't seem to figure it out, which is starting to frustrate me. To see what I mean, go to www.bikescarsandvans.co.uk/test.php and select the drop down menu. And see coding for particular section. Any help will be much appericated, I know a good sleep would probaly help but don't have time for that right now Code: [Select] print " </td> <td colspan='2'> <div id='CollapsiblePanel" . $car_row["id"] . "' class='CollapsiblePanel'> <div class='CollapsiblePanelTab' tabindex='0'><img class='drop' src='images/drop_down.jpg' alt='Go' /></div> <div class='CollapsiblePanelContent'> <p> <div class='title_tabs'>Standard Specification</div> <table class='standard_spec'> <tr> <th> Engine Size: </th> <td> ".$car_row["engine"]." </td> </tr> <tr> <th> BHP: </th> <td> ".$car_row["bhp"]." </td> </tr> <tr> <th> Fuel Type: </th> <td> ".$car_row["fuel_type"]." </td> </tr> <tr> <th> Number of Doors: </th> <td> ".$car_row["no_doors"]." </td> </tr> <tr> <th> Wheels: </th> <td> ".$car_row["wheels"]." </td> </tr> </table> </p> "; //QUERY FOR TITLE START $query_title = "SELECT * FROM extras JOIN car_to_extra ON (car_to_extra.extras_id = extras.id) WHERE car_to_extra.car_id = '{$car_row['id']}' ORDER BY extras.price ASC"; $title_results = mysql_query($query_title) or die ("Error in query: $query_title. ".mysql_error()); $current_heading = ''; while ($title_row = @ mysql_fetch_array($title_results )) { if ($current_heading != $title_row["title"]) { // The heading has changed from before, so print the new heading here. $current_heading = $title_row["title"]; print " <p> <div class='title_tabs'>" . $title_row["title"] . "</div> "; } ?> <a class='data' href='#' onmouseout='hideTooltip()' onmouseover='showTooltip(event,"<?php print "" . $title_row["info"] . "<br/>(£" . $title_row["price"] . ")"; ?>");return false'> <?php print " <img class='extra_img' src=\"". $title_row["img"] ."\" alt='" . $title_row["img_alt"] . "' /></a> </p>"; } //QUERY FOR TITLE END print " </div> </div> <script type='text/javascript'> <!-- var CollapsiblePanel" . $car_row["id"] . " = new Spry.Widget.CollapsiblePanel('CollapsiblePanel" . $car_row["id"] . "', {contentIsOpen:false}); //--> </script> </td> </tr> "; Here is a link to the website I am designing. http://173.167.65.189/singleton/ the username and password is ksingleton at the bottom where the calendar is the query is searching for scheduled events, i am only posting the company_id right now, the problem I am having is I want a scheduled event that continues through the week to stay in line with the one before. But if the schedule before ended and a new one continues it moves the next date company_id up. How can i align the table rows or create an empty table row on all other days if there was a schedule before. here is the code for the calendar. Quote <?php $db = new mysqli(); $setdate=strtotime($_GET['date']); $session=$_GET['session']; $calview=$_GET['calview']; if ($setdate <> "") { $date=$setdate; } else { $date=time(); } $day=date('d',$date); $month=date('m',$date); $year=date('Y',$date); $first_day=mktime(0,0,0,$month,1,$year); $title=date('F',$first_day); $day_of_week = date('D', $first_day); switch($day_of_week) { case "Sun": $blank = 0;break; case "Mon": $blank = 1;break; case "Tue": $blank = 2;break; case "Wed": $blank = 3;break; case "Thu": $blank = 4;break; case "Fri": $blank = 5;break; case "Sat": $blank = 6;break; } $days_in_month = cal_days_in_month(0, $month, $year); echo "<table cellpadding=0 cellspacing=0 border=0>"; echo "<tbody>"; echo "<tr>"; echo "<td><a href=?session=$session&calview=day>Day</a></td>"; echo "<td><a href=?session=$session&calview=week>Week</a></td>"; echo "<td><a href=?session=$session&calview=month>Month</a></td>"; echo "</tr>"; echo "<tr>"; echo "<td colspan=3>"; if ($calview=="month") { echo "<table border=1 width=800>"; echo "<tbody>"; echo "<tr height=25><th colspan=7> $title $year </th></tr>"; echo "<tr height=25><td width=58 align=center>Sunday</td><td width=58 align=center>Monday</td><td width=58 align=center>Tuesday</td><td width=58 align=center>Wednesday</td><td width=58 align=center>Thursday</td><td width=58 align=center>Friday</td><td width=58 align=center>Saturday</td></tr>"; $day_count = 1; echo "<tr height=50>"; while ( $blank > 0 ) { echo "<td></td>"; $blank = $blank-1; $day_count++; } $day_num = 1; while ($day_num <= $days_in_month ) { $newdate = $month ."/". $day_num ."/". $year; $msqlnewdate = date('m/d/Y',strtotime($newdate)); echo "<td valign=top><table cellpadding=0 cellspacing=0 border=0><tr><td><a href=?session=$session&calview=day&date=$newdate> $day_num </a></td></tr>"; $db = new mysqli('localhost','username','password', 'rsdata'); if (!$db) { echo 'ERROR: Could not connect to database.'; } else { $query = $db->query("SELECT * FROM rentals WHERE date_format(str_to_date(pdate_out,'%Y-%m-%d'), '%m/%d/%Y') <= '$msqlnewdate' AND date_format(str_to_date(pdate_in, '%Y-%m-%d'), '%m/%d/%Y') >= '$msqlnewdate'"); if ($query) { while ($result = $query->fetch_object()) { echo "<tr><td>".$result->company_id."</td></tr>"; } } echo "</table>"; } $day_num++; $day_count++; if ($day_count > 7) { echo "</tr><tr>"; $day_count=1; } } while ($day_count >1 && $day_count <=7 ) { echo "<td></td>"; $day_count++; } echo "</tr></table>"; } else if ($calview=="week") { } else { } echo "</td>"; echo "</tr>"; echo "</tbody>"; echo "</table>"; ?> MOD EDIT: DB credentials edited out. Hello all
I'm new to this forum. I've been struggling on this problem for a few days now.
I’m writing a script to automatically push code to a git server.
exec(‘git config user.name "’ . $userName . ‘"’);
exec(‘git config user.email "’ . $userEmail . ‘"’);
exec('git checkout -b ’ . $branch);
exec(‘git add --all’);
exec(‘git commit -m "’ . $message . ‘"’);
exec('git push origin ’ . $branch);
When running the last command, the script stops and asks for a user name, then a password. I tried other forums, searching the net. I'm frustrated...
Thanks in advance. i have a mysql table which contains name like mid mname 101 AAA 102 BBB 103 CCC now i have to print this name in a html table like AAA, BBB, CCC i am getting this by while loop in a variable but when loop changes then value also change so please tell me how i get this only in one variable & print Hello there, i have a pretty simple problem, but i cant seem to get over it hence this post. I have a very simple form with 2 text boxes and a dropdown box. The user is supposed to insert an actor to the database using that form, but the thing is that i can't get the value from my dropdown box for some reason. Anyway here's the form. Code: [Select] <div id="content"> <?php if (isset($_POST["actorID"]) && isset($_POST["actorName"]) && isset($_POST["nation"])) { if ($_POST["actorID"] != "" && $_POST["actorName"] != "") { $link = mysql_connect('localhost', 'Ugluth', 'dracul'); if (!$link) { die('Could not connect: ' . mysql_error()); } mysql_select_db("videoclub", $link); $sql="INSERT INTO actors (Actor_ID, Actor_Name, Actor_Nation) VALUES ('$_POST[actorID]','$_POST[actorName]','$_POST[nation]')"; if (!mysql_query($sql,$link)) { die('Error: ' . mysql_error()); } else { echo "1 record added"; mysql_close($link); } } else { echo "All fields need to be completed to add the record."; } } $link = mysql_connect('localhost', 'turu', 'tururu'); if (!$link) { die('Could not connect: ' . mysql_error()); } mysql_select_db("videoclub", $link); $result_nation = mysql_query("SELECT * FROM nationalities"); $row = mysql_fetch_array($result_nation); mysql_close($link); ?> <form action="actor_add.php" method="post"> <table> <tr> <td> Actor ID: </td> <td> <input type="text" name="actorID"> </td> </tr> <tr> <td> Actor Name: </td> <td> <input type="text" name="actorName"> </td> </tr> <tr> <td> Actor Nationality ID: <?php ?> </td> <td> <select name="nation"> <?php while($row = mysql_fetch_array($result_nation)) { ?> <option value="<?php $row[0]; ?>"><?php echo $row[1]; ?></option> <?php } ?> </select> </td> </tr> </table> <?php mysql_free_result($result_nation); ?> <input type="submit" value="Submit" /> </form> </div> <!-- end #content --> I'm trying to enter the value to my query using '$_POST[nation]' on the $sql string i use for query. I have also tried just displaying it on the screen to see what value it holds, but it just wont show anything. Also i don't really need this answered for this particular problem, but in general its good to know, how can i select the value from the text of each option on the drop down box? The $row[0] holds the id while $row[1] holds the name. Thank you in advance and if u need any more information just let me know Hi I have coded a drop down menu with php and i am trying to retrieve the data when a user select a option from the menu and the data is retrieved from the database. So far i have tried and nothing is displaying when i tried to process the php form. Sales.php Page <form action="saleprocess.php" method="GET"> <?php echo 'Product Model:'; $query="SELECT * FROM products"; /* You can add order by clause to the sql statement if the names are to be displayed in alphabetical order */ $result = mysql_query ($query); echo "<select name=product_model value=Select>Product Model</option>"; // printing the list box select command while($rows=mysql_fetch_array($result)){//Array or records stored in $nt echo "<option name=product_model value='.$rows[product_id].'>$rows[product_model]</option>"; /* Option values are added by looping through the array */ } echo "</select><br>"; ?> <input type='submit' name='submit' value='Create'></input> <br> </form> ******************************************************************************** Salesprocess.php page <?php include("connect.php"); if(isset($_GET['product_id'])){ $product_id = $_GET['product_id']; $query = mysql_query("SELECT * FROM products WHERE product_id= $product_id"); while($rows = mysql_fetch_assoc($query)) { echo 'Product Model<br>'; echo $rows['product_id']; echo $rows['product_model']; } } ?> Muchly appreciated if someone can help me Hi everyone, On my website i have a drop down box on the account information page that has a list of countries. In the example below, im using the php to do 2 things, 1: if the user changes the country and clicks submit but an error is found on the page, rather than revert back to the original country it stays as the newly selected one. 2: if the user does not change the country, the country from the database is simple displayed. As you can see below it is taking up quite a bit of code. I was wondering if anybody has any ideas how i can make this smaller? <select name="country"> <option value="select" selected>Select...</option> <option value="Afganistan" <?php if ($submit) { if ($country == "Afganistan"){ echo "selected";}} else {if ($row['country'] == "Afganistan") {echo "selected";}} ?>>Afghanistan</option> <option value="Albania" <?php if ($submit) {if ($country == "Albania"){echo "selected";}} else {if ($row['country'] == "Albania") {echo "selected";}} ?>>Albania</option> <option value="Algeria" <?php if ($submit) {if ($country == "Algeria"){echo "selected";}} else {if ($row['country'] == "Algeria") {echo "selected";}} ?>>Algeria</option> <option value="American Samoa" <?php if ($submit) {if ($country == "American Samoa"){echo "selected";}} else {if ($row['country'] == "American Samoa") {echo "selected";}} ?>>American Samoa</option> <option value="Andorra"<?php if ($submit) {if ($country == "Andorra"){echo "selected";}} else {if ($row['country'] == "Andorra") {echo "selected";}} ?>>Andorra</option> <option value="Angola"<?php if ($submit) {if ($country == "Angola"){echo "selected";}} else {if ($row['country'] == "Angola") {echo "selected";}} ?>>Angola</option> Thanks Alright so I created a MySQL database that has 5 tables each named a brand of a dirt bike. In each table their are 2 fields, one INDEX_ID and MODELS. Under that for rows I have every model bike named. A quick question before I get onto what I want to do: Can data be added underneath the rows? For example, users will be able to submit information about each bike model. If each bike model is a row, can there be a category past that row or do I need to make each field a model name and just have a ton of fields and have the rows be the information users submit. To make it easier to understand I'll post the SQL code for the brand Honda: CREATE TABLE `Honda` ( `INDEX_ID` int(3) NOT NULL auto_increment, `MODELS` varchar(20) collate latin1_general_ci NOT NULL, PRIMARY KEY (`INDEX_ID`) ) ENGINE=MyISAM DEFAULT CHARSET=latin1 COLLATE=latin1_general_ci AUTO_INCREMENT=23 ; -- -- Dumping data for table `Honda` -- INSERT INTO `Honda` VALUES(1, 'CR85'); INSERT INTO `Honda` VALUES(2, 'CR125'); INSERT INTO `Honda` VALUES(3, 'CR250'); INSERT INTO `Honda` VALUES(4, 'CRF100'); INSERT INTO `Honda` VALUES(5, 'CRF150'); INSERT INTO `Honda` VALUES(6, 'CRF230'); INSERT INTO `Honda` VALUES(7, 'CRF250X'); INSERT INTO `Honda` VALUES(8, 'CRF250R'); INSERT INTO `Honda` VALUES(9, 'CRF450X'); INSERT INTO `Honda` VALUES(10, 'CRF450R'); INSERT INTO `Honda` VALUES(11, 'CRF50'); INSERT INTO `Honda` VALUES(12, 'CRF70'); INSERT INTO `Honda` VALUES(13, 'CRF80'); INSERT INTO `Honda` VALUES(14, 'XR650'); INSERT INTO `Honda` VALUES(15, 'CR500'); INSERT INTO `Honda` VALUES(16, 'XR100'); INSERT INTO `Honda` VALUES(17, 'XR200'); INSERT INTO `Honda` VALUES(18, 'XR250'); INSERT INTO `Honda` VALUES(19, 'XR400'); INSERT INTO `Honda` VALUES(20, 'XR50'); INSERT INTO `Honda` VALUES(21, 'XR70'); INSERT INTO `Honda` VALUES(22, 'XR80'); Anyway I still need to figure out how to have this under a form that a user can use to select the bike they want to submit information about. A code like this perhaps?: <? $connection = mysql_connect("localhost","user","pass"); $fields = mysql_list_fields("dbname", "table", $connection); $columns = mysql_num_fields($fields); echo "<form action=page_to_post_to.php method=POST><select name=Field>"; for ($i = 0; $i < $columns; $i++) { echo "<option value=$i>"; echo mysql_field_name($fields, $i); } echo "</select></form>"; ?> Thanks. Hey guys, I have a drop down menu like so <select name="Categories[]"> <Option value="Horror" Name="Horror">Horror</Option> <Option value="Romance" Name="Romance">Romance</Option> ......................................... Like that, just out of interest once the user has selected the option I am unsure on how to extract what the user chose from the drop down menu, how do you grab the value from array of the drop down menu that the user has selected. Any help A.S.A.P would really be appreciated. It also has to be an array. I also thought of using the $_POST['Categories']; to try grab the value if that is the correct way to go about it. Thank you in advance. Hi guys, basically here pull out the data from database then creating taxt field automatically and submit into anther table. everything works fine but data not inserting in to the table. could you guys check my code please? <?php $con = mysql_connect("localhost","root",""); mysql_select_db("uni", $con)or trigger_error('MySQL error: ' . mysql_error()); ?> <?php $result = mysql_query("SELECT * FROM course") or trigger_error('MySQL error: ' . mysql_error()); echo '<select name ="cid[]">'; while($row = mysql_fetch_array($result)) { echo '<option value="' . $row['CourseID'] . '">'. $row['CourseID'] .'</option>'; } echo '</select>'; //------------------ ?> <?php if(!empty($_POST["submit"])) { $value = empty($_POST['question']) ? 0 : $_POST['question']; ?> <form name="form1" method="post" action="result.php"> <?php for($i=0;$i<$value;$i++) { echo 'Question NO: <input type="text" name="qno[]" size="2" maxlength="2" class="style10"> Enter Marks: <input type="text" name="marks[]" size="3" maxlength="3" class="style10"><br>'; } ?> <label> <br /> <br /> <input type="submit" name="Submit" value="Submit" class="style10"> </label> </form> <?php } else{ ?> <form method="POST" action="#"> <label> <span class="style10">Enter the Number of Question</span> <input name="question" type="text" class="style10" size="2" maxlength="2"> </label> <input name="submit" type="submit" class="style10" value="Submit"> </form> <?php }?> result.php <?php $con = mysql_connect("localhost","root","") or die('Could not connect: ' . mysql_error()); mysql_select_db("uni",$con) or die('Could not connect: ' . mysql_error()); foreach ($_POST['cid'] as $c) {$cid [] = $c;} foreach($_POST['qno'] as $q){$qno[] = $q;} foreach($_POST['marks'] as $m){$marks[] = $m;} $ct = 0; for($i=0;$i<count($qno);$i++) { $sql="INSERT INTO examquesion (CourseID,QuesionNo,MarksAllocated) VALUES('$cid[$i]','$qno[$i]','$marks[$i]')"; mysql_query($sql,$con) or die('Error: ' . mysql_error()); $ct++; } echo "$ct record(s) added"; mysql_close($con) ?> As a complete newbie to php and webdesigning i have a following problem.I would like to retrieve the data from database and display it in a drop down menu.Then i should allow the user to select the values from drop down list along with other details,in other words i have to embed the drop down output as the form input for the user and store the form data in another table.I am running a xampp server and i am using php 5.4 version.Please help.My code is as follows.In this case project_name is displayed as the drop down output.but how do i use the same drop down output as a input in the form. <html> <head></head> <body> <?php error_reporting(E_ALL ^ E_DEPRECATED); include 'connect.php' ; $tbl_name="projects"; $sql="SELECT project_name FROM $tbl_name "; $result=mysql_query($sql); if($result === FALSE) { die(mysql_error()); } ?> <form name="resources" action="hourssubmit.php" method="post" > <?php echo "<select name='project_name'>"; while ($row = mysql_fetch_array($result)) { echo "<option value='" . $row['project_name'] ."'>" . $row['project_name'] ."</option>"; } echo "</select>"; ?> </form> </body> </html> what im trying to do is take a youtube embed code find the URL code for that video and remove all other parts of the code keeping only the URL of the video after i get the URL by it self then replace the http://www.youtube.com/ part of the URL with http://i2.ytimg.com/vi/ to do this i know that i need something like this to get the URL of the video http://www.youtube.com/((?:v|cp)/[A-Za-z0-9\-_=]+) but how to only return just the URL is something idk how to do then use str_replace http://www.youtube.com/(?:v|cp)/" with http://i2.ytimg.com/vi/ so in the end im asking if anyone know how to remove all other codes i dont want and only keep the URL this may have to be done using "regex" im not sure as i dont know to much about regex and what it can do but does sound like it would help lol Dear All Good Day, i am new to PHP (a beautiful server side scripting language). i want to send a mail with line by line message i tried with different types like by placing the things in table and using <br /> but the thing is the tags are also visible in the message of the mail. Here is my code: $message1 = "Name :". $_REQUEST['name']."<br />"; $message1 .= "Surname :". $_REQUEST['surname']."<br />"; $message1 .= "Cellphone :". $_REQUEST['mobileno']."<br />"; $message1 .= "Telephone :". $_REQUEST['landno']."<br />"; $message1 .= "Fax :". $_REQUEST['fax']."<br />"; $message1 .= "Company :". $_REQUEST['company']."<br />"; $message1 .= "Email :". $_REQUEST['email']."<br />"; $message1 .= "Country :". $_REQUEST['country']."<br />"; $message1 .= "Enquity :". $_REQUEST['enquiry']."<br />"; $message1 .= "Date&Time :". $date."<br />"; For this code if try to print/echo it it is working fine it is displaying line by line, but using this variable ($message1) in the mail these <br /> are also visible. Can any one guide me to resolve(to remove these tags from the message part) this issue. Thanks in Advance. :confused: Hi. I want a simple textbox, that when submited, will replace very every new line, with the <br> tag. What will happen, when it submits, it will take the contents of a textbox, add the <br> tag where a new line is suposed to be, and save the string to a MySQL Database. I think this is the easiest way of allowing a user to edit what appears on a website when logged in, but if there is a easier way, then please tell me. What I am trying to do, is a login page, and once logged in, you enter new text into the textbox, click submit, and on the website homepage, the main text will change to what was submitted. But if there is a new line, I think the only way in HTML to make one is to put a <br> tag, so I want the PHP to but that tag wherever there is a new line. Sorry if I am confusing, I am not that advanced at PHP, but I would be very happy if you could supply me with the correct code. Time is running out... If you do not understand me, please tell me -- PHPLeader (not) i have to read a single line from a csv, its a really big file and i only need one column.
i need the response to be a string ,i made a search and found the following code but i dont have any idea how to get a single line from a single string per run .
<?php
$row = 1;
//open the file
if (($handle = fopen("file.csv", "r")) !== FALSE)
{
while (($data = fgetcsv($handle, 0, ",")) !== FALSE)
{
$num = count($data);
echo "<p> $num fields in line $row: <br /></p>\n";
$row++;
for ($c=0; $c < $num; $c++)
{
echo $data[$c] . "<br />\n";
}
}
fclose($handle);
}
?>
Edited by bores_escalovsk, 16 May 2014 - 06:38 PM. I have a script that reads a .gz file into an array and prints the name of each record but will not work on larger files. Is there a way to read 1 line at a time? Here is the code I have so far. Code: [Select] <?php if ($handle = opendir('.')) { print "<ol>"; while (false !== ($file = readdir($handle))) { if($file != '..' && $file!="." && $file!="start_update.php" && $file!="sharons_dbinfo.inc.php" && $file!="root.php" && $file!="read_directory.php" && $file!="read_dir.php" && $file!="new_category.php" && $file!="index.php" && $file!="file_count.php" && $file!="dir_loop2.php" && $file!="dir_loop1.php" && $file!=".htaccess" && $file!="Answer.txt" && $file!="Crucial_Technology-Crucial_US_Product_Catalog_Data_Feed.txt"){ $filename = $file; $go = filesize($filename); if($go >= 1){ $filename2 = explode("-", $filename); $filename2 = $filename2[0]; echo str_replace("_"," ",$filename2) . ' | Filesize is: ' . filesize($filename) . ' bytes<br>'; $gz = gzopen($filename, 'r'); $lines = gzfile($filename,10000); foreach ($lines as $line) { $line2 = explode(",", $line); $line2 = str_replace("," , "-" , $line2); echo "<li>".str_replace("," , "-" , $line2[4])."</li><br>"; } } } } closedir($handle); } ?> </ol> How to get this echo line to display as one line? No matter what I have done it displays as two lines. I even tried <nobr></nobr> Teachers Name: John Jones $userid = mysql_real_escape_string($_GET['user_id']); $sql = "select * from users where `id`='$userid' "; $rs = mysql_query($sql); while($row = mysql_fetch_array($rs)) { echo "<h3>Teachers Name: </h3>" . $row["first_name"] . " " . $row["last_name"] ; } Thanks for your help. I'm so sorry to ask such a "newbee question," but believe me I have been on Google for the better part of the week trying to find an answer.
I'll start with the brief question.
Then I'll give an example to show what I mean.
Then, I'll give the BESTEST "pseudo-code" I could come up with (to PROVE that I've really given it my best).
Question:
How do I make PHP loop through a big file, make the changes line by line, and to save the resultant file.
Example: I have a 100mb text file ("animals.txt") with 500,000 lines. If any lines have the word "cat" in it, I want to add "Be careful with cats!" to the end of the line:
From this:
A fish and his tank.
A cat and his toy.
A bird and her cage.
A frog and his lilly.
A cat and her friend.
To this:
A fish and his tank.
A cat and his toy. Be careful with cats!
A bird and her cage.
A frog and his lilly.
A cat and her friend. Be careful with cats!
The best "pseudo-code" I can come up with is:
<?php
$data = file_get_contents("animals.txt");
$lines = explode(PHP_EOL,$data); I want my application will send a email after 10 minutes of sending another email. In my application A user completes registration with payment Application sends the user a payment confirmation emailNow I want to send another email 10 minutes After payment confirmation email with welcome tipsBelow is the function where for user setup .
public function finishUserSetup($Sub){
if($Sub == 0){
$subscription = SubscriptionPlans::where('identifier', '=', "Monthly")->first();
$expiry = date('Y-m-d', strtotime('+' . $subscription->months . ' months'));
$sub_period = "monthly";
} else{
$subscription = SubscriptionPlans::where('identifier', '=', "Annually")->first();
$expiry = date('Y-m-d', strtotime('+' . $subscription->months . ' months'));
$sub_period = "annually";
}
$this->expiry_date = $expiry;
$this->user_type = "SUB";
$this->subscription_period = $sub_period;
$this->update();
$replaceArray = array(
'fullname' => $this->forename . " " . $this->surname,
'subscriptionName' => $subscription->name,
);
EmailTemplate::findAndSendTemplate("paymentconfirm", $this->email, $this->forename . " " . $this->surname, $replaceArray);
}
In the above function the last line of code is the one which sends a payment confirmation email to the user which is
EmailTemplate::findAndSendTemplate("paymentconfirm", $this->email, $this->forename . " " . $this->surname, $replaceArray);
I want to execute the following line of code 10 minutes after the above one
EmailTemplate::findAndSendTemplate("WelcomeTips", $this->email, $this->forename . " " . $this->surname, $replaceArray);
How to do that. that is running the last line of code 10 minutes after Hi All, I have 2 tables: one CarMake - CarMakeID - CarMakeDesc two CarModel - CarModelID - CarModelMake - CarModelDesc Depending on what the user selects in the first dropdown (carmake) the possible selection in the second dropdown (model) needs to be limited to only the models from the selected carmake. in the second table (Carmodel : the 'CarModelMake' = CarMakeID, to identify the make) How do I limit the dropdown 'CarModel' based on the selected CarMake in the first dropdown. link : http://98.131.37.90/postCar.php code : -- -- -- Code: [Select] <label> <select name="carmake" id="CarMake" class="validate[required]" style="width: 200px;"> <option value="">Select CAR MAKE...</option> <?php while($obj_queryCarMake = mysql_fetch_object($result_queryCarMake)) { ?> <option value="<?php echo $obj_queryCarMake->CarMakeID;?>" <?php if($obj_queryCarMake->CarMakeID == $CarAdCarMake) { echo 'selected="selected"'; } ?> > <?php echo $obj_queryCarMake->CarMakeDesc;?></option> <?php } ?> </select> </label> <label> <select name="carmodel" id="CarModel" class="validate[required]" style="width: 200px;"> <option value="">Select MODEL...</option> <?php while($obj_queryCarModel = mysql_fetch_object($result_queryCarModel)) { ?> <option value="<?php echo $obj_queryCarModel->CarModelID;?>" <?php if($obj_queryCarModel->CarmodelID == $CarAdCarModel) { echo 'selected="selected"'; } ?> > <?php echo $obj_queryCarModel->CarModelDesc;?></option> <?php } ?> </select> </label> I'm a novice.. and appreciates all the help ! |
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