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PHP - Echo Checked
Hello all,
The following script is used to query an array. (www.example.com/test.php?lev[]=MS&lev[]=College) Currently, it echos the values queried and simply lists them (echo "$lev (br />\n"). Is it possible to echo those values by keeping the checkbox in question checked instead? Code: [Select] <?php echo "<b>Levels <br /> />"; if(!empty($_GET['lev'])) foreach($_GET['lev'] as $lev){ echo "$lev <br />\n"; } ?> <form /> <input type="checkbox" name="lev[]" value="PreK" /><font class="sm">PreK</font> <input type="checkbox" name="lev[]" value="Elem" /><font class="sm">Elem</font> <input type="checkbox" name="lev[]" value="MS" /><font class="sm">MS</font> <input type="checkbox" name="lev[]" value="HS" /><font class="sm">HS</font> <input type="checkbox" name="lev[]" value="College" /><font class="sm">College</font> <input type="submit" value=" ..:: Submit ::.. " /> Thank you. ~Wayne Similar TutorialsOK, have no idea what's going on... I've done this a million times... why wont this output!?? I must have a major brain meltdown and dont know it yet!!! Code: [Select] <?php // this echoes just fine: echo $_POST['testfield']; // but this wont echo: echo if (isset($_POST['testfield'])) { $_POST['testfield'] = $test; } echo $test; /// or even this DOESNT echo either!: $_POST['testfield'] = $test; echo $test; ?> Hi All, I'm trying to echo the response from an SLA query, the query works and returns the data when I test it on an SQL application.. but when I run it on my webpage it won't echo the result. Please help? <?php $mysqli = mysqli_connect("removed", "removed", "removed", "removed"); $sql = "SELECT posts.message FROM posts INNER JOIN threads ON posts.pid=threads.firstpost WHERE threads.firstpost='1'"; $result = mysqli_query($mysqli, $sql); echo {$result['message']}; ?> How to write the code to get the count of the word "CHECKED" from a row? I tried this and it's not working. Please help. $query = mysql_query("SELECT field1, field2, field3, COUNT(CHECKED) as counter FROM db1 WHERE row_name = 'science'"); while($line = mysql_fetch_array($query)) { echo $query['counter']; } i am having issues checking which value is checked in a form. here is my code for it: Code: [Select] <input type="radio" name="accountActivation" id="no" value="no" checked=" <? if (isset($register_errors)) { if (($_POST['accountActivation']) == "no") { echo "checked"; } } ?> " /> when i first open the page, no is checked. why is this, when there should be no errors, and the radio should not even be checked in the first place? Hi guys, I wonder if any if you could help me: I have created a login page with "remember me" function... The page will redirect either to Admin Page or User Page depends their 'level'. But when I logged in as Admin and checked the box "remember me" and close all the browser to see IF the "remember me" work. When I open a new browser - and go to the index page - it is okay, I can view the page.... but it redirect me to the USER page NOT Admin page. Does anyone know why this happens? So I need to echo a row from my database with php, but where i need to echo is already inside an echo. This is my part of my code: $con = mysql_connect("$host","$username","$password"); if (!$con) { die('Could not connect: ' . mysql_error()); } mysql_select_db("main", $con); $result = mysql_query("SELECT * FROM Vendor"); while($row = mysql_fetch_array($result)) { //I need to echo right here .................. but I get a blank page when I try this. Please Help. echo '<option value=$row['vendor_id']>'; echo $row['vendor_id']; echo '</option>'; } mysql_close($con); Result: A Blank page. Thanks in advance! I have a log system that allows 10 logs on each side(Left and right). I am trying to make it so that the left side has the 10 most recent logs, then the right as the next 10. Any ideas? This topic has been moved to JavaScript Help. http://www.phpfreaks.com/forums/index.php?topic=318750.0 Hi: How do I check to see if a checkbox has not been checked? Would the same rules apply to a radiobutton? This is what I have: Code: [Select] <input name="status[1]" id="approved" value="<?php echo $_POST['approved'] ?>" type="checkbox" /> <label for="approved">Approved</label> Code: [Select] if (isset($_POST['status[1]'])): $errMsg['status[1]'] = "Must select a status."; endif; I tested it out, but it just prints the error message even if I have checked it I am using the following piece of code to open a new window when a checkbox is checked... <INPUT type="checkbox" name="name" value="YES" onClick="window.open('/somepage.php?do=something&id=<?php echo $SOMEID["id"] ?>','','width=600,height=350,left=100,top=100,screenX=100,screenY=100')"> The page opens ok when the checkbox is ticked (checked). PROBLEM: The page opens again if the checkbox is un-ticked (un-checked). I just want to open the page if the checkbox is ticked only but not sure how to do this? OTHER INFO: Seperate checkboxes are dynamically created on the page for each id=<?php echo $SOMEID["id"] ?> hi all I have 3 checkboxes whose values are getting stored in the database with a comma separator. Code: [Select] <input name="color[]" type="checkbox" value="red" id="red" />red <input name="color[]" type="checkbox" value="blue" id="blue" />blue <input name="color[]" type="checkbox" value="green" id="green" />green values getting stored with comma separator Code: [Select] red, blue, green now i want to show the checkboxes as "checked" to show the customer as which colors they chose lasttime when they submitted the form. How can i make it possible ? If it is possible without the comma separator then i can remove the comma. vineet What is the best way using php to see if a checkbox is checked or not. I see so many different options out there googling, that I am interested knowing the most robust method. Thanks Hello, I'm new to the forum so bare with me. I'm trying to take data entered into a text box on a form and have that data checked against an associative array. I've attached a .doc with some pictures showing how the data should be displayed. It's the checking part that I'm having trouble with. Any help would be appreciated Thanks Hello, Hope someone can help me cause i been at it all day and google woul'dn't help me. I been writing a little script which generates a list of numbers you can check and make a list of the numbers that have been checked. Now when i use $_get to the next page everything works but now i need to take the url and use $_get back to the generated check box list and it needs to know by the url which box had been checked and checked that one so the list can be altered. here is my code sofar Code: [Select] <!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=utf-8" /> <title>AH plaatjes.</title> </head> <body name="top"> Meer info beneden aan de pagina. <a href="#beneden">Naar beneden.</a> <p> <script language=JavaScript> function set(n) { temp = document.form1.elements.length ; for (i=0; i < temp; i++) { document.form1.elements[i].checked=n; } } function checkall() { void(d=document); void(el=d.getElementsByName('check')); for(i=0;i<el.length;i++) void(el[i].checked=1) } //invert function invers(){ temp = document.form1.elements.length ; for (i=0; i < temp; i++){ if(document.form1.elements[i].checked == 1){document.form1.elements[i].checked = 0;} else {document.form1.elements[i].checked = 1} } } </script> </p> <form id="form1" name="form1" method="get" action="./readlist.php"> <label>Naam eigenaar lijstje:<input name="naam" type="text" /></label> <label>Lijst van selectie plaatjes<select name="collect"> <option value="0">selecteer optie!</option> <option value="1">in decollectie</option> <option value="2">nog niet in collectie</option> <option value="3">dubbele collectie</option> </select></label> <p> <?php //TO Do trying to get the checkboxes checked with get and array ///////////////// do { $i = $i+1; echo " <input type=\"checkbox\" name=\"checked[]\" id=\"check\" value=\"". $i ."\"/> <label for=\"". $i ."\">". $i ."</label> <br /> "; if($i > 324){break;} }// end do while($i < 325); ?> </p><input name="check" type="submit" value="Print lijst" /> <input type=button onClick="checkall()" value="selecteer alles" name="button"> <input name=button onclick="invers()" type=button value=" Keer selectie om "> <INPUT name=button onclick=set(0) type=button value=" Reset "> </form> <p><a name="beneden">De geselecteerde nummers komen overzichtelijk in een lijstje. Zo heb je heel overzichtelijk een lijst met de nummers die je al hebt of juist nog niet hebt. Die kan je dan doorsturen selecteren en kopieren met je rechtermuis knop en naar je kennissen of vrienden sturen.</a> ook kan je de url (webadres) kopieren en doorgeven.</p> <p><a href="#top">Naar boven.</a></p> </body> </html> Now what i need is when you fill out an url like http://localhost/ah/index.php?checked%5B%5D=2&checked%5B%5D=3 it needs to check the same boxes. please help me out thanks in advance while ($div=mysql_fetch_array($query4, MYSQL_ASSOC)) { //$div - shows all the entries $row22=mysql_fetch_array($query2, MYSQL_ASSOC); //row22 - shows all the entries that should be checked $name= array($div['date_id']); $datess=$row22['date_id']; if (in_array($datess, $name, true)) { echo '<input type="checkbox" value="' . $div[date_id] .'" checked>' . $div[date] . '</option>'; }else{ echo '<input type="checkbox" value="' . $div[date_id] .'">' . $div[date] . '</option>'; } 1. when echo $row22 shows only the entries that should be checked - working good 2. checkboxes show checked if only the first checkbox is selected or all 3 - if a combination is selected, it is all blank. Any ideas? Thank you Hey, I'm new to this stuff. But I'm calling a Table with a list of products, I want people to be able to compare the products they checked. what is the best way of doing this with the code below? I put "blah" for everything except for the compare table and calls. This code seemed to work the best for the look I wanted for it. They all call the checkbox from 'Compare' in the mysql table. Thanks! $result = mysql_query("SELECT * from Compare_Tool ORDER BY Blah ASC"); //Table starting tag and header cells echo "<table border='1'><tr><th>Compare</th><th>blah</th><th>blah</th><th>blah</th><th>blah</th><th>blah</th><th>blah</th><th>blah</th><th>blah</th><th>blah</th><th>blah</th><th>blah</th></tr>"; while($row = mysql_fetch_array($result)){ ?> <tr> <td align="center"><input name="checkbox[]" type="checkbox" Compare="checkbox[]" value="<? echo $row['Compare']; ?>"></td> <td ><? echo $row['blah']; ?></td> <td ><? echo $row['blah']; ?></td> <td ><? echo $row['blah']; ?></td> <td ><? echo $row['blah']; ?></td> <td ><? echo $row['blah']; ?></td> <td ><? echo $row['blah']; ?></td> <td ><? echo $row['blah']; ?></td> <td ><? echo $row['blah]; ?></td> <td ><? echo $row['blah]; ?></td> <td ><? echo $row['blah]; ?></td> <td ><? echo $row['blah]; ?></td> </tr> <?php } echo "</table>"; ?> <?php $goal = rand(1, 3); ?> <td><div align="left"> Goal</div></td> <td><label> <div align="center"> <input type="radio" name="goal_chk" id="radio2" value="1" <?php if($goal == "1") { echo "checked"; }?> /> </div> </label></td> <td><label> <div align="center"> <input name="goal_chk" type="radio" id="radio13" value="2" <?php if($goal == "2") { echo "checked"; }?> /> </div> </label></td> <td><label> <div align="center"> <input type="radio" name="goal_chk" id="radio23" value="3" <?php if($goal == "3") { echo "checked"; }?> /> </div> </label></td> </tr> i have a group of radio buttons whose value is checked randomly everytime you refresh the page, now that works fine, everytime you refresh the page the checked values moves randomly but the problem i have is getting the actual of a checked radio button. i have tried this... $goal_checked = $_POST['goal_chk']; echo $goal_checked; But nothing happens. Any ideas on how to make this possible. Thanks a lot you guys. This topic has been moved to JavaScript Help. http://www.phpfreaks.com/forums/index.php?topic=358408.0 This topic has been moved to JavaScript Help. http://www.phpfreaks.com/forums/index.php?topic=349419.0 Hi all, I have been trying to figure this out for a few days already and can't seem to figure it out. I want to display a few images with a checkbox next to it so I can delete multiple images at once when I press the delete button. My problem is that I can't seem to figure out how to check if the checkbox is checked. if I uncheck one of them and click the delete button it deletes one from the array but I don't know which one was unchecked. For example if I uncheck the second one the array shows: Array ( => on [1] => on ) but I don't know which of the three was unchecked? Here is the code I have been testing with: Code: [Select] <?php // if submit button is clicked if(isset($_POST['submit'])) { // do delete function(); } // display deletethis[] array if (isset($_POST['deletethis'])) { echo "<pre>"; print_r ($_POST['deletethis']); echo "</pre>"; } $separate = array("http://funnypicturesimages.com/images/image/funny-dog-pictures.jpg", "http://1.bp.blogspot.com/-35wQMpYtNZc/TXWNx8y2xCI/AAAAAAAB_2o/9vZYNfWrGn8/s400/funny_demotivational_posters_01.jpg", "http://3.bp.blogspot.com/-TFnzZ8zFtgg/TXWNpodBkGI/AAAAAAAB_2Q/O_fOOSqFM6w/s400/funny_demotivational_posters_04.jpg"); echo '<form action="'.$_SERVER['PHP_SELF'].'" method="post">'; // display each image from array foreach ($separate as $value) { echo "<img src=".$value.">"; if ($_POST['deletethis'] = "on") { $checked="checked"; } else { $checked=""; } if ($_POST['deletethis'] != "") { echo "checked"; } else{ echo "unchecked"; } // if checkbox currently checked display it checked else display unchecked if ($checked == "checked") { echo '<input type="checkbox" name="deletethis[]" checked="checked"/><br /><br />'; } else { echo '<input type="checkbox" name="deletethis[]"/><br /><br />'; } } ?> <center><input type="submit" name="submit" value="Delete Checked"></center> </form> |
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