PHP - Php/ajax <select> Innerhtml In Ie
Hi,
It appears I ran into the dreaded IE <select> issue with innerhtml. I have looked at multiple solutions but alas I reach out for help! The code works beautifully in Safari and Firefox but fails in IE by simply loading blanks when AJAX is called. I read up on microsoft.com a DIV may work outside the <select>. But I must be messing something up. Thanks for your puzzle solving help! Code: AJAX: xmlHttp.onreadystatechange=function(){ if(xmlHttp.readyState==4) { var temp=(xmlHttp.responseText); document.getElementById("option").innerHTML=temp; } } var queryString = "?page=" + page; xmlHttp.open("GET","AJAX_page.php" + queryString,true); xmlHttp.send(null); PHP: <?php echo "<form action='next.php' method='POST'>"; echo "<select name='id' id='option'><option value=0>Option_0/option>"; { echo "<option value ='$id'>$choice</option>"; } echo "</select>"; echo "<input type='submit' value='Submit'></form>"; ?> Similar TutorialsThis topic has been moved to Ajax Help. http://www.phpfreaks.com/forums/index.php?topic=347691.0 I had this working, but when I try and get fancy and use AJAX the data doesn't display. I think this is a PHP problem though. My code for the select form including AJAX code Code: [Select] <!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Strict//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-strict.dtd"> <html xmlns="http://www.w3.org/1999/xhtml" xml:lang="en-GB"> <head> <title>AJAX Example</title> <link rel="stylesheet" type="text/css" href="Form.css" media="screen" /> <script type="text/javascript" src="http://ajax.googleapis.com/ajax/libs/jquery/1.4.2/jquery.min.js"></script> <script type="text/javascript"> $(document).ready(function(){ $("tr:odd").addClass("odd"); }); </script> <script type="text/javascript"> function showPlayers(str) { var xmlhttp; if (str=="") { document.getElementById("DataDisplay").innerHTML=""; return; } if (window.XMLHttpRequest) {// code for IE7+, Firefox, Chrome, Opera, Safari xmlhttp=new XMLHttpRequest(); else {// code for IE6, IE5 } xmlhttp=new ActiveXObject("Microsoft.XMLHTTP"); } xmlhttp.onreadystatechange=function() { if (xmlhttp.readyState==4 && xmlhttp.status==200) { document.getElementById("DataDisplay").innerHTML=xmlhttp.responseText; } } xmlhttp.open("GET","Query.php?category_id="+str,true); xmlhttp.send(); } </script> </head> <body> <h1">AJAX Example</h1> <?php #connect to MySQL $conn = @mysql_connect( "localhost","username","pw") or die( "You did not successfully connect to the DB!" ); #select the specified database $rs = @mysql_SELECT_DB ("MyDB", $conn ) or die ( "Error connecting to the database test!"); ?> <form name="sports" id="sports"> <legend>Select a Sport</legend> <select name="category_id" onChange="showPlayers(this.value)"> <option value="">Select a Sport:</option> <?php $sql = "SELECT category_id, sport FROM sports ". "ORDER BY sport"; $rs = mysql_query($sql); while($row = mysql_fetch_array($rs)) { echo "<option value=\"".$row['category_id']."\">".$row['sport']."</option>\n "; } ?> </select> </form> <br /> <div id="DataDisplay"></div> </body> </html> Query.php <?php #get the id $id=$_GET["category_id"]; #connect to MySQL $conn = @mysql_connect( "localhost","username","pw") or die( "Error connecting to MySQL" ); #select the specified database $rs = @mysql_SELECT_DB ("MyDB", $conn ) or die ( "Could not select that particular Database"); #$id="category_id"; #create the query $sql ="SELECT * FROM sports INNER JOIN players ON sports.category_id = players.category_id WHERE players.category_id = '".$id."'"; echo $sql; #execute the query $rs = mysql_query($sql,$conn); #start the table code echo "<table><tr><th>Category ID</th><th>Sport</th><th>First Name</th><th>Last Name</th></tr>"; #write the data while( $row = mysql_fetch_array( $rs) ) { echo ("<tr><td>"); echo ($row["category_id"] ); echo ("</td>"); echo ("<td>"); echo ($row["sport"]); echo ("</td>"); echo ("<td>"); echo ($row["first_name"]); echo ("</td>"); echo ("<td>"); echo ($row["last_name"]); echo ("</td></tr>"); } echo "</tr></table>"; mysql_close($conn); ?> I think the problem is either with this part in the AJAX Code: [Select] xmlhttp.open("GET","Query.php?category_id="+str,true); or most likely in my Query.php code when I wasn't using AJAX and using POST it worked fine, but adding the AJAX stuff and GET it doesn't work. When I echo out the SQL the result is Code: [Select] SELECT * FROM sports INNER JOIN players ON sports.category_id = players.category_id WHERE players.category_id = '' so the category_id is not being selected properly and that is the primary key/foreign key in the MySQL table which connects the JOIN. In my registration form there are many form elements to give users' details for registration in my web site. among those elements there are two select boxes for user to select their district and city. I have created these two select box using ajax. Therefor a user select a district then automatically ajax creating second select box for cities is populating. I used separate php page called findcity.php to create city select box. I called this findcity.php page from my original register.php page through onChange attribute. and there I passed the district id with the url to findcity.php page. like wise, Now I need to bring city id to my original register.php page when user select a city from city select box in findcity.php page. my problem is that. I tried to get city Id to register.php page but still I couldn't get it. city id is needed me to send to the database with other form elements' values. can anybody help me to fix my problem? here is my coding for your reference. This code is, from my register.php page Code: [Select] <div> <label for="district">District <img src="../images/required_star.png" alt="required" /> : </label> <?php require_once ('../includes/config.inc.php'); require_once( MYSQL2 ); $query="select * from district order by district_id"; $result = mysqli_query( $dbc, $query); echo '<select name="district" class="text" onChange="getCity(' . "'" . 'findcity.php?district=' . "'" . '+this.value)">'; echo '<option value="">-- Select District --</option>'; while( $row = mysqli_fetch_array($result, MYSQLI_NUM)) { echo '<option value="' . $row[0] . '"'; // Check for stickyness: if ( isset( $_POST['district']) && ( $_POST['district'] == $row[0] )) echo ' selected="selected"'; echo " >$row[1]</option>"; } echo '</select>'; ?> </div> <div> <label for="city">City <img src="../images/required_star.png" alt="required" /> : </label> <input type="hidden" name="reg_locationid" id="reg_locationid" value="56" /> <div id="citydiv" style="position: relative; top: -14px; left: 130px; margin-bottom: -26px;"> <select name="city" class="text"> <option>-- Select City --</option> </select> </div> </div> this is, from my findcity.php page Code: [Select] <?php $districtId=$_GET['district']; require_once ('../includes/configaration.inc.php'); require_once( MYSQLCONNECTION ); $query="select city_id, city_name from city2 where district_id=$districtId"; $result=mysqli_query( $dbc, $query); echo '<select name="city" class="text"> <option>-- Select City --</option>'; while($row=mysqli_fetch_array($result, MYSQLI_NUM)) { echo '<option value="' . $row[0] . '"'; // Check for stickyness: if ( isset( $_POST['city']) && ( $_POST['city'] == $row[0] )) { echo ' selected="selected"'; //echo '<input type="hidden" name="city" value="' . $row[0] . '"'; } echo " >$row[1]</option>"; } echo '</select>'; ?> 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. 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 It has been ten years since I last wrote any code. I am trying to use PHP to write into a HTML element. I know JavaScript has the innerHTML function but I cannot seem to find any equivalent PHP function. My questions are 1) Is there an equivalent PHP function to innerHTML? If not 2) Is there an intelligent method of using PHP fire JavaScript innerHTML Why? I want to get data from MySQL DB using PHP and print one record at a time to the webpage. I will be using a <div> tag as the data will be longer then one line so I do not want to use a span tag. I would like to force line breaks therefore I am using CSS with 'display:block;' I have not found a way to print the data from MySQL How to set innerHTML of DomNode? I am parsing html.I want to change a html of DomNode/DomElement? How to achieve the same? Thanks in advance CSJakharia hirealimo.com.au/code1.php this works as i want it: Quote SELECT * FROM price INNER JOIN vehicle USING (vehicleID) WHERE vehicle.passengers >= 1 AND price.townID = 1 AND price.eventID = 1 but apparelty selecting * is not a good thing???? but if I do this: Quote SELECT priceID, price FROM price INNER JOIN vehicle....etc it works but i lose the info from the vehicle table. but how do i make this work: Quote SELECT priceID, price, type, description, passengers FROM price INNER JOIN vehicle....etc so that i am specifiying which colums from which tables to query?? thanks Hello, i am trying to pull the innerHTML out of this: Code: [Select] <a href="(.*?)">(.*?)</a> here is what I have: Code: [Select] <?php $html = file_get_contents("http://www.businessinvestingsource.com/blcheck2.html"); preg_match_all('/<a href="(.*?)">(.*?)<\/a>/', $html, $links, PREG_SET_ORDER); foreach ($links as $link) { $linkto = $link[1]; $anchor = $link[0]; echo "<b>Link:</b> ".$linkto."<br /><b>Anchor:</b> ".$anchor."<br /><br /> "; } ?> Now this code works but the innerHTML is coming out as a link I want it to come out as plaintext you can view he http://businessinvestingsource.com/anchorcheck2.php Can anyone help? Thank you. what is the best way yo do this? Code: [Select] <b><font color=green>Get me out please </font> . </b> thanks in advance As the title say, I can not for the life of me get the "$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 { $bank = 'You must choose a bank before continuing'; } ?> change-content.js Code: [Select] window.onload = init; // finds all <select> tags will class="select-content" and activates function function init() { var sel = document.getElementsByTagName("select"); for (var i=0; i<sel.length; i++){ if (sel[i].className == "select-content") { sel[i].onchange = getFile; } sel[i].selectedIndex = 0; } } function getFile (url) { var url = "AJAX/get_Bank.php?dealType="+ this.value; if (window.XMLHttpRequest) {xmlhttp=new XMLHttpRequest();} else {xmlhttp=new ActiveXObject("Microsoft.XMLHTTP");} xmlhttp.open("GET",url,false); xmlhttp.send(); // a loop that looks through all <div>s on the page // and then replaces the id with the value and gets that file var divs = document.getElementsByTagName("div"); for (var i=0; i<divs.length; i++) { if(divs[i].id == "bank") { divs[i].id = this.value; divs[i].innerHTML=xmlhttp.responseText; } } } I have 2 queries that I want to join together to make one row
Dear All, I wish to have 2 drop down boxes, Country Select Box and Locality Select Box. The locality select box will be affected by the value chosen in the country select box. All is working fine except that the locality select box is not being populated. I know that the problem is in the sql statement WHERE country_id='$co' because i am having an error that $co is an undefined variable. All the rest works fine because i have replaced the $co variable directly with a number (say 98) for a particular country id and it worked fine. In what way can i define this variable $co so that it is accepted by my sql statement? Thank you for your help in advance. MySQL Tables indicated below: CREATE TABLE countries( country_id INT(3) UNSIGNED NOT NULL AUTO_INCREMENT, country_name VARCHAR(30) NOT NULL, PRIMARY KEY(country_id), UNIQUE KEY(country_name), INDEX(country_id), INDEX(country_name)) ENGINE=MyISAM; CREATE TABLE localities( locality_id INT(10) UNSIGNED NOT NULL AUTO_INCREMENT, country_id INT(3) UNSIGNED NOT NULL, locality_name VARCHAR(50), PRIMARY KEY (locality_id), INDEX (country_id), INDEX (locality_name)) ENGINE=MyISAM; Extract PHP script included below: // connect to database require_once(MYSQL); if(isset($_POST['submitted'])) { // trim the incoming data /* this line runs every element in $_POST through the trim() function, and assigns the returned result to the new $trimmed array */ $trimmed=array_map('trim',$_POST); // clean the data $co=mysqli_real_escape_string($dbc,$trimmed['country']); $lc=mysqli_real_escape_string($dbc,$trimmed['locality']); } ?> <form action="form.php" method="post"> <p>Country <select name="country"> <option>Select Country</option> <?php $q="SELECT country_id, country_name FROM countries"; $r=mysqli_query($dbc,$q) or trigger_error("Query: $q\n<br />MySQL Error: " . mysqli_error($dbc)); while($row=mysqli_fetch_array($r)) { $country_id=$row[0]; $country_name=$row[1]; echo '<option value="' . $country_id . '"'; if(isset($trimmed['country']) && ($trimmed['country']==$country_id)) echo 'selected="selected"'; echo '>' . $country_name . '</option>\n'; } ?> </select> </p> <p>Locality <select name="locality"> <option>Select Locality</option> <?php $ql="SELECT locality_id, country_id, locality_name FROM localities WHERE country_id='$co' ORDER BY locality_name"; $rl=mysqli_query($dbc,$ql) or trigger_error("Query: $q\n<br />MySQL Error: " . mysqli_error($dbc)); while($row=mysqli_fetch_array($rl)) { $locality_id=$row[0]; $country_id=$row[1]; $locality_name=$row[2]; echo '<option value="' . $locality_id . '"'; if(isset($trimmed['locality']) && ($trimmed['locality']==$locality_id)) echo 'selected="selected"'; echo '>' . $locality_name . '</option>\n'; } // close database connection mysqli_close($dbc); ?> </select> </p> <p><input type="submit" name="submit" value="Submit" /></p> <input type="hidden" name="submitted" value="TRUE" /> </form> Just thought I would ask as I see that the AJAX help board is something of a ghost town (much like the MS SQL board). Since my post there doas also have some php in it, and I am in no way certain my problem is not in the php could anyone that knows a bit of both please have a look at it he http://www.phpfreaks.com/forums/index.php?topic=343419.0 cheers, just hope this circumvents the double posting rule i have a select box and need to change the data with the selection.it is working correctly , my problem is that in the data i have a paging .when we going to next page no data showing . ie select box will not have any value. how to solve this problem. thanks in advance How should I structure XMLHTTPRequests? Should I create a controller called Ajax and post/get all ajax calls through the methods of that controller? What have other people done in the past? What worked best for you? Hi guys, I am after some simple code/tutorial that shows how to make an AJAX/jQuery function to fetch data from a PHP MySQL Database on a timed interval. So far I have not been able to find a simple example of this so if anyone can help out that would be very much appreciated. Cheers in advance. Pardon my noobness, but I'm learning to wrap AJAX into my work and use it to get XML instead of "static" PHP that generates the HTML. The login/security portion has my head spinning, but it's probably not as difficult as I think and I'm probably just confusing myself. In the past, for each PHP page in my site, I would perform a quick salted login check based on the username/password stored in the $_SESSION variables. Perhaps it was a bit overboard to check on each page, but, well, I did it. With AJAX, I *NEED* to ensure that the php resulting from an AJAX POST request won't run if the user isn't authenticated, and I need to ensure that they didn't just somehow force a $_SESSION variable to reflect an authenticated session. I also need to ensure that someone can't just load up the PHP page on it's own, somehow send a POST to it and run it without being authenticated. I suppose that beyond the larger picture of "How do I ensure that the user is authenticated, the POST request is authentic, and nobody has forced a change in the $_SESSION stored on the server, I have a few specific questions. I know that in part I'm confused about the whole cookie/SESSION process. In my old PHP site, the SESSION number was stored on the cookie on the user's machine. If the info is sent via AJAX, does the PHP get the SESSION info from the cookie or does it have to be explicitly sent? With potentially several users sending AJAX requests at the same time, how will my PHP know which SESSION to use for each request? Is is secure enough to set an "Autheticated" flag in $_SESSION once the user is authenticated the first time? Is it really just as simple as sending a username/salted password hash as AJAX/POST and setting an authenticated flag in the SESSION to ensure that the rest of the AJAX application runs without allowing someone to back-door the PHP? |