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PHP - Help With Forming Php Echo Ajax Link
Hi all,
I am having problems in forming this link: { echo "<a title='page $i' class='current' href='javascript:ajaxpage('view.php?catid=" . $catid ."&" ; if($subid > 0) { echo "subid="; echo $subid . "&"; } echo "p=$i', 'content';'>$i</a> "; } Which currently points to javascript:ajaxpage( and then ends! Similar TutorialsHi, This is my first post in this forum and am a PHP beginner. I have written a script in php and need to use echo to see the values of variables etc. However I dont get the output on the screen while using echo . I am using Xampp server with Apache. The startup.html file code is
<!DOCTYPE HTML>
</head>
and the test.php code is <?php
$lat = $_REQUEST['latitude'];
echo 'latitude- '.$lat . ', longitude- ' . $lon; Please help regards
Sanjish Hey, $upgrade_next_sql = "SELECT * FROM todo_upgrades WHERE profile_id = ".$profile_id." AND level = ".$profile_data['level_'.$show.'']+1." AND type = '$show'"; doesn't work because there is "+1". How can I add this value to the variable without forming a extra variable before? How do I make this output a clickable link: <?php echo $_POST["website"]; ?><br /> I have created a database that echos out the result on a webpage, what i want to do is create one of the echo results into a link - the code is as follows: "Web Address: $web_address <br>" . Is there an easy way to make the result a link as the input will always be a web address? Thanks in advance!
Hi guys,
echo " Download <a href=".$file_path." download>Here</a>"; ?>
Didnt work, just took me to the path of the image. What i want is a link and when clicked the download window pops out.
Thanks What I'm trying to do is take two arrays and combine them using array_combine. After they are combined I want to take each key and value pair and use them in a string to build a MySQL query. See below for some pseudo-code =D <?php $fields = array('first_name', 'last_name'); //the fields that will be set with an UPDATE query. $newvals = array('Joe', 'Blow');//The values that will go into $fields $arr = array_combine($fields, $newvals);//Each value now has a key of the field it will update. /** This is where I want to return my SET string with something like 'SET ' . $arr['first_name'] ' = ' . $arr['first_name_value'] ', etc etc. */ There it is, I suppose. Are there any easier ways of doing what I'm trying to accomplish, that is, am I on the right track or just making things harder for myself? Both arrays will have dynamic values throughout my application, so I need to be able to get both each field and value for each query. Any links to some constructs, etc, etc? That's all I really need and I can post the solution when I've figured it out. hello, im trying to add a hyperlink that launches in a new window to the following (in the last column) any ideas? Code: [Select] echo "<tr>"; echo "<td align='center'>" . $row["ID"] . "</td>"; echo "<td align='center'>" . $row["Name"] . "</td>"; echo "<td align='center'>" . $row["jobNO"] . "</td>"; echo "<td align='center'>" . $epn . "</td>"; echo "<td align='center'>" . $cname . "</td>"; echo "<td align='center'>" . $cadd . "</td>"; //want to add a hyperlink here echo "</tr>"; } echo "</table>"; } I have to put a link that can be clicked around the first name and last name variables in the echo statement below. I also need to add a variable from the Select query to the link such as: 'www.phpfreaks.com/admin/customers.php?page=1&cID="VARIABLE HERE" &action=edit' I've tried a bunch of things and none seem to work. while($row = mysql_fetch_array($result)){ echo '<tr><td>' . $row['customers_firstname'] . " " . $row['customers_lastname'] . '</td><td>' . date("M. d, Y", strtotime($row['date'])) . '</td><td>' . $row['plan_id'] . '</td></tr>'; David Trying to figure out how to make it so name is a link to the profile when its echo anyone know how to do this? im at a huge stand still Code: [Select] <?php $sql = "SELECT name FROM users WHERE DATE_SUB(NOW(),INTERVAL 5 MINUTE) <= lastactive ORDER BY id ASC"; $query = mysql_query($sql) or die(mysql_error()); $count = mysql_num_rows($query); $i = 1; while($row = mysql_fetch_object($query)) { $online_name = htmlspecialchars($row->name); echo '<a href="Inbox.php">"'[$goauld]'</a>"'; ?> <html> <?php include connect.inc; $id = $_GET['id']; mysql_query("UPDATE count SET clicks=clicks+1 WHERE id='$id'"); $sql = mysql_query("SELECT link FROM count WHERE id='$id'"); $fetch = mysql_fetch_row($sql); $result = mysql_query($sql); while($row = mysql_fetch_array($result)) { echo "Name :{$row['id']} <br>" . "Subject : {$row['link']} <br>" . "Message : {$row['count']} <br><br>"; } mysql_close(); ?> <a href='/index.php?id=1'>link1</a> <a href='/index.php?id=2'>link2</a> </html> Hi, I would like to send a value in a link and request it back in ajax and set it as a variable in php in the same page, can anyone help me here? thanks OK, 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; ?> Right now I redirect to index page after I delete a record. However I am looking to make it so that I can delete a record without redirecting the page. I know this can be accomplised using Ajax. I have spent countless hours before trying to make it work, but it did not work.
So here is a basic setup I created. Can you please update it with ajax code so that I can see how it's done properly?
<!DOCTYPE HTML> <html lang="en"> <head> <meta charset="UTF-8"> <title>Home Page</title> </head> <body> <div class="record" > <a href="record.php?id=<?php echo $record_id ?>"><?php echo $record_name; ?></a> <div class="delete-record"> <a href="delete.php">Delete Record</a> </div> </div> </body> </html> Edited by man5, 18 August 2014 - 08:55 PM. 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']}; ?> Now I'm having this strange issue with my website I'm currently working on a tester system and I've encountered a problem that I'm unable to find the issue, tho I'm thinking my ajax php part of the script to be the thing causing it even tho it seems strange that it would cause it. The first part which is connected to where the problem occurs is the echo"<form>"; and from there, It should take you to index.php?page=tester&select=answer, now that is where it in the browser goes there tho it still shows the page stuff from the last page which is index.php?page=tester&select=applications, so it's like showing both &select=answer and &select=applications on the same page. <?php $q=$_GET["q"]; include'../config/connection.php'; $result = mysql_query("SELECT * FROM applications WHERE id = '$q'"); echo "<center><table border='1'> <tr> <th>Account Name</th> <th>Character Name</th> <th>Gender</th> <th>Skin Color</th> </tr>"; $row = mysql_fetch_array($result); echo "<tr>"; echo "<td>" . $row['name'] . "</td>"; echo "<td>" . $row['charactername'] . "</td>"; echo "<td>" . $row['gender'] . "</td>"; echo "<td>" . $row['race'] . "</td>"; echo "</tr></table></center>"; echo"<br/>"; echo"<table><tr> <th>Description</th> <th>Metagaming</th> <th>Powergaming</th></tr>"; echo"<tr>"; echo "<td><textarea readonly='readonly' style='width:22em; height:20em;'>".$row['description']."</textarea></td>"; echo "<td><textarea readonly='readonly' style='width:22em; height:20em;'>".$row['mg']."</textarea></td>"; echo "<td><textarea readonly='readonly' style='width:22em; height:20em;'>".$row['pg']."</textarea></td>"; echo"</tr></table><table><br/><center><h1>Answer</h1><br/><form action='index.php?page=tester&select=answer' method='post'>"; echo"<textarea name='why' style='height:10em; width:60em;'></textarea><br/>"; echo"<input type='submit' name='answer' value='Accept' /><a/>"; echo"<input type='submit' name='answer' value='Decline' /></center>"; echo"<input type='hidden' name='id' value='$q'/>"; echo"</form></table>"; ?> Now on &select=answer it included a page which the script of that include consist of the stuff below, it outputs that the query was successfully, and all that. <? if(!empty($_POST['why'])) { $why = mysql_real_escape_string($_POST['why']); $answer = trim($_POST['answer']); $id = $_POST['id']; if($answer == "Accept") { $query1 = mysql_query("UPDATE characters SET accepted = '1' WHERE id = '".$id."'"); echo"Successfully accepted"; $answer = 1; } elseif($answer == "Decline") { echo"Successfully declined"; $answer = 0; } $query = mysql_query("UPDATE applications SET answer = '$why' AND tester = '".$_COOKIE['Username']."' AND accepted = '$answer' AND answered = '1' WHERE cid = '".$id."'") or die('Could not connect: ' . mysql_error()); if($query) { echo"<br/>Query went through without problems"; header("Refresh: 5;url=index.php?page=tester"); } } ?> This is the ajax part javascript of it which gets the information for index.php?page=tester&select=applications Code: [Select] <script type="text/javascript"> function showApplication(str) { if (str==""||str==0) { document.getElementById("txtHint").innerHTML=""; return; } if (window.XMLHttpRequest) {// code for IE7+, Firefox, Chrome, Opera, Safari xmlhttp=new XMLHttpRequest(); } xmlhttp.onreadystatechange=function() { if (xmlhttp.readyState==4 && xmlhttp.status==200) { document.getElementById("txtHint").innerHTML=xmlhttp.responseText; } } xmlhttp.open("GET","tester/applications.php?q="+str,true); xmlhttp.send(); } </script>If you need any more information feel free to ask for it. Thanks in advance. Hello, I've been trying this for hours now, looking at different examples and trying to change them to work for me, but with no luck... This is what I am trying to do: I have a simple form with: - 1 input field, where I can enter a number - 1 Submit Button When I enter a number into the field and click submit, I want that number to be send to the php file that is in the ajax call, then the script will take that number and run a bunch of queries and then return a new number. I want that new number to be used to call the php script via ajax again, until no number is returned, or something else is returned like the word "done" or something like that, at which point is simply makes an alert or populated a div with a message... The point is, that depending on the number entered it could take up to an hour to complete ALL the queries, so I want the script that is called to only run a fixed amount of queries at a time and then return the number it is currently at (+1), so that it can continue with the next number when it is called again. I would like to use jquery, but could also be any other way, as long as I get this to work. I already have the php script completed that needs to be called by the ajax, it returns a single number when being called. Thank you, vb 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? As the title say, I can not for the life of me get the "$bank" content to display, no matter HOW much I try... Does anyone see any errors. I am sooooooo wiped out at this! main page <? $body = ' <script type="text/javascript" src="change-content.js"></script> <div id="addSold"> <form action="'.$_SERVER['REQUEST_URI'].'" method="post" name="form" autocomplete="off"> <fieldset id="Vehicle"> <legend>Vehicle</legend> <ul> <li><label for="Year">Year</label>'.$Year.'</li> <li><label for="Make">Make</label>'.$Make.'</li> <li><label for="Model">Model</label>'.$Model.'</li> <li><label for="Trim">Trim</label><input type="text" name="Trim" id="Trim" size="10" value="'.$trim.'" disabled="disabled"></li> </ul> <ul> <li><label for="Mileage">Mileage</label><input type="text" name="Mileage" id="Mileage" size="5" maxlength="6" value="'.$row['mileage'].'"></li> <li><label for="VIN">VIN</label><input type="text" name="VIN" id="VIN" size="23" maxlength="17" value="'.$row['vin'].'" disabled="disabled"></li> <li><label for="Color">Color</label>'.$Exterior.'</li> </ul> </fieldset> <fieldset id="Deal"> <legend>Deal</legend> <ul> <li> <label for="soldDte1">Date</label> <input type="text" name="soldDte1" id="soldDte1" size="1" maxlength="2" onkeyup="return autoTab(this, 2, event)" value="08"> / <input type="text" name="soldDte2" id="soldDte2" size="1" maxlength="2" onkeyup="return autoTab(this, 2, event)" value="30"> / <input type="text" name="soldDte3" id="soldDte3" size="1" maxlength="2" value="'.$year.'"> <a href="#"><img id="date_'.$row[stock].'" src="images/Icons/dateOff.png" onfocus="this.select();lcs(this)" onmouseover="MM_swapImage(\'date_'.$row[stock].'\',\'\',\'images/Icons/dateOn.png\',1)" onmouseout="MM_swapImgRestore()" alt="Choose Date"></a> </li> <li> <label for="salesman">Salesman</label> <select name="salesman" id="salesman"> <option></option> '.$salesmen.' </select> </li> </ul> <ul> <li> <label for="dealType">Deal Type</label> <select name="dealType" class="select-content" onchange="getFile(this.value)"> <option></option> <option value="AL">Auto Loan</option> <option value="Cash">Cash</option> <option value="CAC">Credit Acceptance</option> <option value="IH">In House</option> <option value="SAL">Sensible Auto</option> </select> </li> <li> <label for="tradeDrop">Trade</label> <select name="tradein" id="tradeDrop" onchange="show_hide_trade(this.value);"> <option value="No">No</option> <option value="Yes">Yes</option> </select> </li> </ul> </fieldset> <div id="Bank" class="view">'.$bank.'</div> </form> </div> '; ?> get_Bank.php <? if ($_GET['dealType'] == "AL") { $bank = ' <fieldset id="AL"> <legend>Auto Loan Figures</legend> <ul> <li><label for="price">Price</label><input type="text" name="price" id="price" class="price" size="7" onchange="currency(this)"></li> <li><label for="down">Down</label><input type="text" name="down" id="down" class="price" size="6" onchange="currency(this)"></li> <li><label for="tax">Tax</label><input type="text" name="tax" id="tax" class="price" size="6" onchange="currency(this)"></li> <li><label for="reg">Plates</label><input type="text" name="reg" id="reg" class="price" size="4" onchange="currency(this)"></li> <li><label for="gap">Gap</label><input type="text" name="gap" id="gap" class="price" size="4" onchange="currency(this)"></li> </ul> <ul> <li> <label for="pymtNum">--------------- Payment ---------------</label> <input type="text" name="pymtNum" id="pymtNum" size="3" maxlength="3" onkeyup="return autoTab(this, 3, event)"> @ <input type="text" name="pymtAmnt" id="pymtAmnt" class="price" size="5" onchange="currency(this)"> per <select name="pymtType"> <option value="Weekly" selected="selected">Week</option> <option value="Monthly">Month</option> </select> </li> <li><label for="APR">APR</label><input type="text" name="APR" id="APR" class="rate" size="6" value="19.00"></li> </ul> </fieldset> '; } elseif ($_GET['dealType'] == "CAC") { $bank = ' Credit Acceptance stuff goes here '; } else { $bank = 'You must choose a bank before continuing'; } ?> change-content.js Code: [Select] window.onload = init; // finds all <select> tags will class="select-content" and activates function function init() { var sel = document.getElementsByTagName("select"); for (var i=0; i<sel.length; i++){ if (sel[i].className == "select-content") { sel[i].onchange = getFile; } sel[i].selectedIndex = 0; } } function getFile (url) { var url = "AJAX/get_Bank.php?dealType="+ this.value; if (window.XMLHttpRequest) {xmlhttp=new XMLHttpRequest();} else {xmlhttp=new ActiveXObject("Microsoft.XMLHTTP");} xmlhttp.open("GET",url,false); xmlhttp.send(); // a loop that looks through all <div>s on the page // and then replaces the id with the value and gets that file var divs = document.getElementsByTagName("div"); for (var i=0; i<divs.length; i++) { if(divs[i].id == "bank") { divs[i].id = this.value; divs[i].innerHTML=xmlhttp.responseText; } } } Hi
I try to echo out random lines of a html file and want after submit password to whole content of the same html file. I have two Problems.
1st Problem When I echo out the random lines of the html file I don't get just the text but the code of the html file as well. I don't want that. I just want the text. How to do that?
for($x = 1;$x<=40;$x++) {
$lines = file("$filename.html");
echo $lines[rand(0, count($lines)-1)]."<br>";
}
I tried instead of "file("$filename.html");" "readfile("$filename.html");" But then I get the random lines plus the whole content. Is there anything else I can use instead of file so that I get the random lines of text without the html code?P.S file_get_contents doesn't work either have tried that one.
2nd Problem:
As you could see in my first problem I have a file called $filename.html. After I submit the value of a password I want the whole content. But it is like the program did forget what $filename.html is. How can I make the program remember what $filename.html is? Or with other words how to get the whole content of the html file?
My code:
if($_POST['submitPasswordIT']){
if ($_POST['passIT']== $password ){
$my_file = file_get_contents("$filename.html");
echo $my_file;
}
else{
echo "You entered wrong password";
}
}
If the password isn't correct I get: You entered wrong password. If the password is correct I get nothing. I probably need to create a path to the file "$filename.html", but I don't know exactly how to do that.
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