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PHP - Html Displaying In Textarea?
When I add some text to the front end of the textarea it displays
Code: [Select] bla <br/> bla <br/> bla <br/> How can I stop it from showing the html tags? Thanks. echo'<textarea id="info" style="width:80%; height:60px;">'; } echo $profile['info']; if($_SESSION['user_id'] == $id) { echo'</textarea><input type="submit" style="width:70%" value="send" onclick="updateInterests(document.getElementById(\'info\').value)" />'; Similar TutorialsHi All, New to PHP so any help appreciated
Can anyone explain why this;
<tr><td><label for="notes">Notes:</label> Displays data from a MySQL database (field $notes, VARCHAR(250) ) and this;
<tr><td class=textboxlabel><label for="notes">Notes:</label> displays nothing????
Not sure how much info to provide........ Edited November 20, 2019 by cyberRobotAdded a more meaningful post title I have a textarea displaying data from a mysql table. However, a user discovered today that if they use a carriage return in the textarea, the data comes out with "rn" at the location in the text where the carriage return goes. I have been searching the forums for a while this afternoon and have found quite a few references to this, but have been unable to fix my problem. I am using mysql_real_escape_string before inserting into database (I am not using anything else at insertion time). Is there something else that I should be using at either the insertion into the database or when I display the data? I have tried various combinations of htmlspecialchars() and htmlentities() and nl2br() with no luck. Here is a sample of the output in my textarea: Quote copy diagonal lines and a square with no more than 1/4" overlap or gap at point of closure(modified).rn2.Patient will draw a person ........ Should look like: Quote copy diagonal lines and a square with no more than 1/4" overlap or gap at point of closure(modified). 2.Patient will draw a person ........ Thanks for the help. Matt Hi I have a text area, that I want to display info pulled from a database. I can get the data to show, But can get each entry of the table to display on it's own line. Example: ob1ob2ob3ob4ob5 Should be: ob1 ob2 ob3 ob4 ob5 CODE: Code: [Select] <textarea id="interest" onfocus="clearInterest()" class="textareacss" style="height:212px;overflow:auto;"><?PHP $newInterestSub = Admin_interests_sub::find_by_cat_id($id); foreach($newInterestSub as $newInterestSubs){ echo $newInterestSubs->interest_sub.'\n'; } ?></textarea> Any help would be great. If the html td changes size, the code below should resize the textarea to fit the html td. The code works for old browsers but why is it not working for new browsers?
$(document).ready(function() {
$('textarea').parent().resize(function() {
var $t = $(this);
$t.find('textarea').height($t.height());
}).resize();
});
Hello dear friends, If i've database table (my_table) with one field (text) and i've created form with textarea Code: [Select] <form action="#"> <textarea name="text" id="text"></textarea> <input type="submit" name="Submit" value="Submit"> </form> I can add any text inside it and save it in my database table but when i add html codes it also store it perfect but only 2 codes is not the closing of form and textarea </form> and </textarea> it not stored and automatically not saved Why ! i mean if i paste inside the above form - textarea the following code for example Code: [Select] <form> <textarea>blah</textarea> </form> it will store it as Code: [Select] <form> <textarea>blah and will not store </form> and </textarea> how then i can add html codes with textarea and form inside it without conflict thanks Hi Guys, I have a textarea on a html page with XXX amount of lines. How can I in php insert a separate mysqlrow for EACH line on the html textarea? Thanks G hi, I have some HTML to edit in my database, in my back end administration I have it in a textarea but when I go to edit it, it all messes up and lots of '/' area added, see below: Code: [Select] Site Design by: <a href=\\\\\\\"http://www.jbiddulph.com\\\\\\\" target=\\\\\\\"_blank\\\\\\\" title=\\\\\\\"John Biddulph - Web Development\\\\\\\">jbiddulph.com</a> php code Code: [Select] <p> <label>Site Design by</label> <textarea class="text-input small-input" name="SiteDesignby"><?php echo $row1['SiteDesignby'] ?></textarea> </p> someone suggested adding this to my page: SiteDesignby.value = SiteDesignby.value.replace(/\\*."/g,'"'); I added the code and nothing changed! Can anyone help please? Hi all!!! I could really use some help. I'm burning my brain out trying to figure out the easiest way to go about this. Here is what I am trying to do (in a nutshell) so I could really use full coding examples. I need to create a text area with simple formatting tools such as Bold, Italics, Underline, Bullets, and Numbering. After formatting the text with only those available formatting tools, everything in the text area needs to be converted into html tags. After the conversion, the HTML data that is create is submitted into a single field into a SQL database. Any help will be greatly appreciated!!! I have a form that generates html code, and displays it in a textarea with submit button. I want to submit button to take the data entered in that textarea1 on page 1, and upload to textarea2 on page 2 How can I do this? I have no idea where to start. I am trying to get this to work in an html page. Code: [Select] <head> <?php require_once('fn.inc.php'); $out=output_rss_tag($ret); ?> </head> <head> <title> </title> <style type="text/css"> <!-- h1 {text-align:center; font-family:Arial, Helvetica, Sans-Serif; } p {text-indent:20px; } --> </style> </head> <body bgcolor = "#ffffcc" text = "#000000"> <h1>server status!<?php echo $ret[SONGTITLE]; ?></h1> </body> </html>I cant seem to figure out why its not diplaying the results Many thanks
$('input#name-submit').on('click', function() {
var name = $('input#name').val();
if ($.trim(name) != '') {
$.post('play2.php', {name: name}, function(data) {
$('div#name-data').text(data);
});
}
});
data contains html code but when it is displayed it is displayed as text and shows the html, how can I get the html code to function properly?thanks Hello So in my index.php code i include('somefile.html');. In the html file itself there is an <img src="/images/picture.jpg">. When looking at index.php on a browser, the html looks fine everything loaded fine except for the <img> There is a little broken image link displayed on the page. i checked through the browser if i mistyped the img src link but somefile.html in the browser displays the image fine. But if i try to display the img in the html file through the php file, its a broken link. Any reason why this is happening? Am I missing something here? Thanks in advance. I am retrieving data from a third party's API using AJAX method. I would like to do two things. 1. Display the data records on a page. 2. Create a pagination on the page to display records more efficiently. I am expecting to retrieve possibly hundreds of records. I can normally do this using PHP but since I am retrieving the records using AJAX, the pagination is gonna be a challenge. First things first is to display the records. Here is my AJAX code. It displays the log data fine. But it doesn't display any data in the "output" div. And it gives this error in the console. Uncaught SyntaxError: Unexpected token o in JSON at position 1 at JSON.parse (<anonymous>) at Object.success
<div class="output"></div>
<script>
$.ajax ({
type: 'GET',
url: "https://3rdpartywebsite.com/api/GetCustomers",
dataType: 'json',
processData: false,
contentType: false,
success: function(data) {
console.log(data)
$newData = JSON.parse(data);
$.each($newData, function(i, v) {
$('.output').append(v.LastName);
$('.output').append(v.FirstName);
});
}
});
</script>
What am I doing wrong? Edited November 14, 2020 by imgroootapparently I am doing this wrong. I want my string results from a form that searches DB content to appear within HTML table, tr, td tags. I get the results fine, but the HTML part isn't appearing. How should I be doing this? <?php $string = ''; $result = mysql_query($sql); /// This is the execution if (mysql_num_rows($result) > 0){ while($row = mysql_fetch_object($result)){ echo "<table>"; echo "<tr>"; $string .= "<td>".$row->last_name."</td> "; $string .= "<td>".$row->first_name."</td>"; $string .= "<td>".$row->employee_id."</td>"; $string .= "<td>".$row->title."</b>"; $string .= "<td>".$row->territory."</td>"; $string .= "<td>".$row->district."</td>"; $string .= "<td>".$row->Phase1A_Score."</td>"; $string .= "<td>".$row->Phase1B_Score."</td>"; $string .= "<td>".$row->Phase1_Average."</td>"; $string .= "<td>".$row->Phase1A_HS_Exam."</td>"; $string .= "<td>".$row->Phase1A_HS_Exam_RT."</td>"; $string .= "<td>".$row->Phase1B_HS_Exam."</td>"; $string .= "<td>".$row->Phase1B_HS_Exam_RT."</td>"; $string .= "<td>".$row->Class_Date."</td>"; $string .= "<td>".$row->Awards."</td>"; $string .= "<br/>\n"; echo "</tr>"; echo "</table>"; } }else{ $string = "No matches found!"; } echo $string; ?> Hi everyone I just getting back to PHP after a break but have forgot how to build a HTML table row. Here is the part of the code that builds up rows for a html table... while($rows = mysql_fetch_array($qry)){ $table .= " <tr> <td><input type=\"checkbox\" name=\"C1\" value=\"ON\"></td> <td width=\"87\">$rows['ref']</td> <td width=\"178\">$rows['transactionReferenceNumber']</td> <td>$rows['totalAmountReceived']</td> <td>$rows['myItemQuantity']</td> <td>$rows['flag_1']</td> <td>$rows['flag2']</td> </tr> "; } This causes an error as you might expect. Please tell me the best way to place the html code into $table Hi there,
I'm a bit of a noobie, and I have a class which will allow me to send an email in html.
However when I send an email with a link such as <a href="http://www.google.com">link</a> using the html() function
The email is sent and everything is displayed corrected.
However the link isn't clickable.
Here is the code for the class.
<?
class eMail
{
var $to = array();
var $cc = array();
var $bcc = array();
var $attachment = array();
var $boundary = "";
var $header = "";
var $subject = "";
var $body = "";
function eMail($name,$mail)
{
$this->boundary = md5(uniqid(time()));
$this->header .= "From: $name <$mail>\n";
}
function to($mail)
{
$this->to[] = $mail;
}
function cc($mail)
{
$this->cc[] = $mail;
}
function bcc($mail)
{
$this->bcc[] = $mail;
}
function attachment($file)
{
$this->attachment[] = $file;
}
function subject($subject)
{
$this->subject = $subject;
}
function text($text)
{
$this->body = "Content-Type: text/plain; charset=ISO-8859-1\n";
$this->body .= "Content-Transfer-Encoding: 8bit\n\n";
$this->body .= $text."\n";
}
function html($html)
{
$this->body = "Content-Type: text/html; charset=ISO-8859-1\n";
$this->body .= "Content-Transfer-Encoding: quoted-printable\n\n";
$this->body .= "<html><body>\n".$html."\n</body></html>\n";
}
function send()
{
// CC
$max = count($this->cc);
if($max>0)
{
$this->header .= "Cc: ".$this->cc[0];
for($i=1;$i<$max;$i++)
{
$this->header .= ", ".$this->cc[$i];
}
$this->header .= "\n";
}
// BCC
$max = count($this->bcc);
if($max>0)
{
$this->header .= "Bcc: ".$this->bcc[0];
for($i=1;$i<$max;$i++)
{
$this->header .= ", ".$this->bcc[$i];
}
$this->header .= "\n";
}
$this->header .= "MIME-Version: 1.0\n";
$this->header .= "Content-Type: multipart/mixed; boundary=$this->boundary\n\n";
$this->header .= "This is a multi-part message in MIME format\n";
$this->header .= "--$this->boundary\n";
$this->header .= $this->body;
// Attachment
$max = count($this->attachment);
if($max>0)
{
for($i=0;$i<$max;$i++)
{
$file = fread(fopen($this->attachment[$i], "r"), filesize($this->attachment[$i]));
$this->header .= "--".$this->boundary."\n";
$this->header .= "Content-Type: application/x-zip-compressed; name=".$this->attachment[$i]."\n";
$this->header .= "Content-Transfer-Encoding: base64\n";
$this->header .= "Content-Disposition: attachment; filename=".$this->attachment[$i]."\n\n";
$this->header .= chunk_split(base64_encode($file))."\n";
$file = "";
}
}
$this->header .= "--".$this->boundary."--\n\n";
foreach($this->to as $mail)
{
mail($mail,$this->subject,"",$this->header);
}
}
}
?>
So I have a jobs database with the following columns: id, jobtext, jobdate, and id. This is how it looks right now: http://prahan.com/jobs/display.html.php I have another table called author. In the authorid column in need the results of this query, SELECT name FROM author WHERE id = (SELECT authorid FROM job) , to be displayed for each row. I also want to be able to customize the header title for each column. Thanks in advance! Hello I'm making a guestbook in PHP and I'm having trouble displaying the posts. The user fills a form and sends it, it goes into the database (mysql) and I can then fetch the data. So far so good. I'm then going to use HTML to display the posts to the user. It's code looking like this: Code: [Select] <p> <br /> ID: ---replaceid--- <br /><br /> Time: ---replacetime--- <br /> From: <a href="---replacehomepage---">---replacefrom---</a> <br /> Email: ---replacemail--- <br /> <br /> Comment: ---replacecomment--- <br /><hr> </p> I want to use str_replace in PHP to replace the values in HTML... like this: Code: [Select] $numquery = mysql_query("SELECT * FROM guestbook", $dbconnect); $num = mysql_num_rows($numquery); $i=1; while ($i < $num) { header('Content-type: text/html'); $html = file_get_contents("test.html"); $query = mysql_query("SELECT * FROM guestbook WHERE id='$i'", $dbconnect); $rows = mysql_fetch_row($query); echo str_replace('---replaceid---', nl2br($rows[0]), $html); echo str_replace('---replacetime---', nl2br($rows[1]), $html); echo str_replace('---replacefrom---', nl2br($rows[2]), $html); echo str_replace('---replacemail---', nl2br($rows[3]), $html); echo str_replace('---replacehomepage---', nl2br($rows[4]), $html); echo str_replace('---replacecomment---', nl2br($rows[5]), $html); $i++; } But this only replaces ---replaceid---, leaves the rest and outputs it. Then, only ---replacetime--- is replaced and so on. I hope you understand. It means that 1 guestbook entry is displayed like 6, with only 1 string replaced at each one. What do you think I should do? Worth saying is that I'm not allowed to mix PHP-code with HTML-code Not really sure how to get the images I have stored in MySQL into a html form. I can call-up the text fields from the database but it cannot seem to find the index for the images. Here is my code:- <?php session_start(); mysql_connect("localhost","root","abc") or die ("Error! Cannot connect to database"); mysql_select_db("theimageworks") or die ("Cannot find database"); $query = "SELECT * FROM jobs"; $result = mysql_query($query) or die (mysql_error()); ?> <?php //display data in html table echo "<table>"; echo "<tr><td>Username</td><td align='center'>Message</td><td>Product Image</td></tr>"; while($row = mysql_fetch_array($result)) { echo "</td><td>"; echo $row['username']; echo "</td><td>"; echo $row['message']; echo "</td></tr>"; echo $row['image']; } echo "</table>"; ?> The error message I get is "Notice: Undefined index: image in....." Thanks in advance! Hi, I'm trying to make a dynamic html table to contain the mysql data that is generated via php. I'm trying to display a user's friends in a table of two columns and however many rows, but can't seem to figure out what is needed to make this work. Here's my code as it stands: Code: [Select] <?php //Begin mysql query $sql = "SELECT * FROM friends WHERE username = '{$_GET['username']}' AND status = 'Active' ORDER BY friends_with ASC"; $result = mysql_query($sql); $count = mysql_num_rows($result); $sql_2 = "SELECT * FROM friends WHERE friends_with = '{$_GET['username']}' AND status = 'Active' ORDER BY username ASC"; $result_2 = mysql_query($sql_2); $count_2 = mysql_num_rows($result_2); while ($row = mysql_fetch_array($result)) { echo $row["friendswith"] . "<br>"; } while ($row_2 = mysql_fetch_array($result_2)) { echo $row_2["username"] . "<br>"; } ?> The above simply outputs all records of a user's friends (their usernames) in alphabetical order. The question of how I'd generate a new row each time a certain amount of columns have been met, however, is beyond me. Anyone know of any helpful resources that may solve my problem? Thanks in advance =) |
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