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PHP - If Statement
Hi All,
I have an SQL query that returns a date (31/01/2011 - for example) and this get's placed into a variable - $req_date. I want to be able to display an image when the required date is within 5 days of the current date. Normally in an if statement I would use: <?php if ($ticket_no == 100) { echo "This is ticket 100" } else { echo "This is not ticket 100" ?> However I am struggling to work out the difference between the 2 dates. Any ideas? Cheers Matt Similar TutorialsI'm having issues with the following: Code: [Select] <?php session_start(); $_SESSION['username']=$_POST['username']; $_SESSION['password']=$_POST['password']; if($_SESSION['username']=="username" && $_SESSION['password']=="password"){ if($_GET['product']=="add"){ $content.=' <p><label>Product Name:</label> <input type="text" name="product_name" size="30" /> <label>Product Price:</label> <input type="text" name="product_price" size="5" /> </p> <p><label>Product Category:</label> <input type="text" name="product_category" size="30" /></p> <p><label>Product Link:</label> <input type="text" name="product_link" size="30" /></p> <p><label>Product Image:</label> <input type="text" name="product_image" size="30" /></p> <p><label>Product Tag:</label> <input type="text" name="product_tag" size="30" /></p> <p><label>Product Keywords:</label> <input type="text" name="keyword" size="30" /></p> <p><label>Product Features:</label><br /> <textarea name="product_features" rows="10" cols="60"></textarea> </p> <p><label>Product Pros:</label><br /> <textarea name="product_pros" rows="5" cols="30"></textarea> </p> <p><label>Product Cons:</label><br /> <textarea name="product_cons" rows="5" cols="30"></textarea> </p> <p><label>Product Description:</label><br /> <textarea name="product_description" rows="10" cols="60"></textarea> </p> <p><label>Product Notes:</label><br /> <textarea name="product_notes" rows="5" cols="30"></textarea> </p> '; $logout='<div><a href="./acp_admincp.php?log-out">Log-Out</a></div>'; } elseif($_GET['product']=="view"){ } else{ $content.=' <a href="./admincp.php?product=add">Add New Product</a> <br /> <a href="./admincp.php?product=view">View Products</a> '; } } elseif(isset($_GET['log-out'])){ session_start(); session_unset(); session_destroy(); header("Location: ./admincp.php"); } else{ $content=' <form action="./admincp.php" method="post"> <p><label>Username:</label> <input type="text" name="username" size="30" />'; $content.='</p> <p><label>Password:</label> <input type="password" name="password" /></p>'; $content.='<p><input type="submit" value="Submit" name="Submit" /></p> </form>'; } ?> <!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.1//EN" "http://www.w3.org/TR/xhtml11/DTD/xhtml11.dtd"> <html xmlns="http://www.w3.org/1999/xhtml" xml:lang="en"> <head> <base href="http://ghosthuntersportal.com/" /> <title>Ghost Hunter's Portal - Admin Control Panel</title> <meta http-equiv="content-type" content="text/html; charset=utf-8" /> <meta name="verify-v1" content="" /> <meta name="keywords" content="ghost, hunters, hunter, ghosts, spirit, spirits, paranormal, investigation, investigator, investigators, k2, emf, meter, kii" /> <meta name="description" content="Ghost Hunters Potal. Parnormal research equipment store." /> <meta name="author" content="Andrew McCarrick" /> <meta name="robots" content="index, follow" <link rel="stylesheet" type="text/css" href="style.css" /> </head> <body> <img src="./logo.png" alt="Ghost Hunter's Portal Admin Control Panel" /> <br /> <div style="color: #AA44AA; font-size: 26px; margin-top: -30px; margin-left: 125px;">Admin Control Panel</div> <?php echo $logout; echo $content; ?> </body> </html> I can log-in, and get to the page with the two links on it. However, once I click one of the links it falls back to the log-in page, and it ends up being a never ending loop. It's doing this: Log-In --> Page with links ---> Log-In page again Should be doing this: Log-In --> Page with links --> Add Product page or View Products page I can never get into the the actual sub page. Just to be clear, the address bar actually shows product=add or product=view, but it still shows the log-in page. Hello Everyone, I have to change the if statements to a switch statement, in this game. Can anyone help? Thanks.
var checkKeyPressed = function (e) {
// Press key A or key D to make dog run to the left or right
// The running speed is specified by the step variable, 10 by default.
// If you replace 10 with a larger integer, the dog will run faster.
//press A, dog runs left
if (e.keyCode == "65") {
if (prex[0] > 5) {
prex[0] = prex[0] - step;
}
}
//press D, dog runs right
if (e.keyCode == "68") {
if (prex[0] < right) {
prex[0] = prex[0] + step;
}
}
};
Can someone write me some simple code to check to see if the url has the word "soundcloud" in them. This is what I want to do.... This is my code now... Code: [Select] echo " <h3> $title - $author</h3> </br> <iframe width='640' height='360' src='$link' frameborder='0' allowfullscreen></iframe> </br> </br> $message </br> <font color='#8b8d9b' size='1'>Posted By: $username $date</font> </br></br></br></br></br> "; However I want to pull the link value out of the database and see if it contains the word "soundcloud" and if it does do this...(note the height is changed) Code: [Select] echo " <h3> $title - $author</h3> </br> <iframe width='640' height='100' src='$link' frameborder='0' allowfullscreen></iframe> </br> </br> $message </br> <font color='#8b8d9b' size='1'>Posted By: $username $date</font> </br></br></br></br></br> "; Any help would be greatly appreciated. I'm having a problem whit the pdo - sql statement: It dosen't return anything but when i try $sql = " SELECT en FROM word WHERE MATCH (sp) AGAINST (:word IN BOOLEAN MODE ) "; without the pdo it's working perfectly Code: [Select] $sql = " SELECT en FROM word WHERE MATCH (:sp) AGAINST (:word IN BOOLEAN MODE ) "; $st = $con->prepare($sql); $st->bindValue(":word",$word,PDO::PARAM_STR); $st->bindValue(':sp','sp',PDO::PARAM_STR); $st->execute(); Code: [Select] <?php mysql_connect("localhost","root") or die(mysql_error()); mysql_select_db("Regis") or die(mysql_error()); $m=$_POST['message']; if (isset($_POST['submit'])) { // makes sure they filled it in // checks it against the database $check = mysql_query("SELECT * FROM registration WHERE username = '".$_POST['username']."'")or die(mysql_error()); //Gives error if user dosen't exist $check2 = mysql_num_rows($check); if ($check2 == 0) { echo("<SCRIPT LANGUAGE='JavaScript'>window.alert('That user does not exist in our database')</SCRIPT>") ?> <html><head></head><body> <form name="register" method="post" action="login.php" onsubmit="return a()"> <table border='0'> <input type="hidden" name="message" value="<?php echo $m;?>"> <tr><td>Username:</td><td><input type="text" name="username" maxlength="60"></td></tr> <tr><td>Password:</td><td><input type="password" name="pass" maxlength="10"></td></tr> <tr><td>Confirm Password:</td><td><input type="password" name="pass2" maxlength="10"></td></tr> <tr><th colspan=2><input type="submit" name="sub" value="Register"></th></tr></table> </form> </body> </html> <? while($info = mysql_fetch_array( $check )) { $_POST['pass'] = stripslashes($_POST['pass']); $info['password'] = stripslashes($info['password']); $_POST['pass'] = md5($_POST['pass']); } } } else { ?> <html> <head> <script type="text/javascript"> function addsmiley(code) { var pretext = document.smile.message.value; this.code = code; document.smile.message.value = pretext + code; } function a() { var x = document.smile.message.value; if(x=="") { alert("Please insert an message!"); return false; } } </script> <style type="text/css"> body{ background-color: #d8da3d } </style> </head> <body> <form name="smile" method="post" action="registration.php" onSubmit="return a()" > <input type="hidden" name="message" value="<?php echo $m;?>"> Your message:<br><textarea name='message' cols='40' rows='2'></textarea><br> <img src="smile.gif" alt=":)" onClick="addsmiley(':)')" style="cursor:pointer;border:0" /> <img src="blush.gif" alt=":)" onClick="addsmiley('*blush*')" style="cursor:pointer;border:0" /> <input type='submit' name='submit' value='Set Name' class='biasa' ></form> <br> <br> </body> </html> <?php } ?> im doing if statments if the $check2==0 it will show the registration form and else it will show the form name="smile" method="post" action="registration.php" onSubmit="return a()" > <input type="hidden" name="message" value="<?php echo $m;?>"> Your message:<br><textarea name='message' cols='40' rows='2'></textarea><br> <img src="smile.gif" alt="" onClick="addsmiley('')" style="cursor:pointer;border:0" /> <img src="blush.gif" alt="" onClick="addsmiley('*blush*')" style="cursor:pointer;border:0" /> <input type='submit' name='submit' value='Set Name' class='biasa' ></form> but my problem is it cannot direct to else matter although the name is already in database may i know which error is wrong or the if statement is wrong Good morning,
I am trying to return specific values depending on the value of the WJCStatusID, for example if WJCStatusID = < 4 but > 2 return "In production"
SQL QUERY:
SELECT tblWJCItem.AddedDescription,
tblWJC.WJCPrefix + Convert(Varchar(10),tblWJC.WJCNo) AS OurRef,
tblWJCItem.MaterialName, tblStockFamily.StockFamily,
tblWJCItem.WeightToSend,
tblWJC.DateCreated,
tblWJC.WJCStatusID
FROM tblWJC
INNER JOIN tblWJCItem ON tblWJC.WJCID = tblWJCItem.WJCID
INNER JOIN tblStockFamily ON tblWJCItem.ProductFamilyID = tblStockFamily.StockFamilyID
IF (tblWJC.WJCStatusID) < 2 THEN 'Pre Production'
ELSE IF (tblWJC.WJCStatusID) < 4 THEN 'In Production'
ELSE IF (tblWJC.WJCStatusID) > 4 THEN 'Ready To Ship'
ELSE 'Awaiting Lab Results';
I am quite new to the world of mssql so I maybe doing something wrong which is quite simply to fix. Can any body help?Hi there can anybody help me please? my if statement isnt working?? All i am trying to do is display an image if the session variable is == entered varialble "Empire" I get the error message "Parse error: syntax error, unexpected T_VARIABLE, expecting " Code: [Select] if $_SESSION['MM_Username'] == 'Empire'{ echo'<img src="Images/empicon.jpg" width="20" height="20" class="moncalimari"/>' ; } ?> I have checked and the session does == Empire Please help Thank you. im trying to make it so before it inserts the comment and username into the database it check if the username is active: 0 or 1. if it is 0 then die but if active then its all good lol. what i have now lets anybody comment. to me it looks perfect ? =[ Code: [Select] if (isset($_POST['submit'])) { $check = mysql_query("SELECT active FROM users WHERE active ='1'") or die(mysql_error()); $check2 = mysql_num_rows($check); if ($check2 != 1) { die('You are Not allowed to comment untill your account is activated.'); }else{ $comment = mysql_real_escape_string(stripslashes(trim($_POST['comment']))); $insert = "INSERT INTO homecomments (username, comment) VALUES ('[$username]', '[$comment]')"; $add_member = mysql_query($insert); { echo "<META HTTP-EQUIV=\"Refresh\" CONTENT=\"1; URL=index.php\">"; } } } I have a simple if statment issue. Here is the code <?php $address = $User['address_l1']; if (empty($address)) { echo "You have not entered an address";} else {echo "You Have entered an address";} ?> The variable $address in this case is not empty so it should echo you HAVE entered an address but its echoing NOT. Can somebody please explain why this is not working? I'm just trying something new out here and wanted to know how it looked AND I'm having a hard time figuring out how to accomplish this with the else statement. Obviously on the else it would mean that there was already a template with the same name existing in the database table so I want the user to be sent back to the original form and I'll implement a jquery dialog box telling the user that there is already an existing template with that name. if (isset($_POST['edittemplate'])) { $templatename = mysqli_real_escape_string($dbc, $_POST['templatename']); $headercode = mysqli_real_escape_string($dbc, $_POST['headercode']); $footercode = mysqli_real_escape_string($dbc, $_POST['footercode']); $templateID = (int)$_POST['templateID']; $query = "SELECT * FROM `templates` WHERE `templatename` = $templatename"; $result = mysqli_query ( $dbc, $query ); // Run The Query $rows = mysqli_num_rows($result); if ($rows == 0) { $query = "UPDATE `templates` SET `templatename` = '$templatename', `headercode`='".$headercode."', `footercode`='".$footercode."' WHERE `id` = '$templateID'"; mysqli_query($dbc,$query); } else { } } So I am trying to tweak a few things on a page and I am stuck on this one. I want the ###### to also be the data in $row['id'] Code: [Select] <?php if (file_exists('/var/www/html/images/races/######/images/' . $row['id'] . $dthumb)): ?> so if $row['id'] = 124 it would be looking for /var/www/html/images/races/124/images/124.jpg')) I know it has to be that I am stumped on the correct question to ask or search for. Any help is appreciated. Thanks! I need help with my if/else statement. Code: [Select] function show_list() { $query = "SELECT show_id,show_name,host,description_short,show_graphic1 FROM shows"; $result = mysql_query($query) or die(mysql_error()); echo "<table>"; while($row = mysql_fetch_array($result)) { echo "<tr> <td rowspan='3' width='155' class='show_image'> <a href='". $site ."show_page.php?show_id=". $row['show_id'] ."'> <img src='". $site ."images/". $row['show_graphic1'] ."' height='150' width='150'></img></a> </td> <td class='show_name'> <h1>"; if ($row['host'] == "") { echo $row['show_name']" with " $row['host']; } else { echo $row['show_name']; } echo " </h1> </td> </tr> <tr> <td height='100'>" . $row['description_short'] . "</td> </tr> <tr> <td align='right'> <a href='". $site ."show_page.php?show_id=". $row['show_id'] ."'>More Info</a> </td> </tr>"; } echo "</table>"; } This function is for a page showing a list of shows from a database. Not all shows have a host. So it does not need to show "with host name" for those shows. I tried if / else statement stating that if there "is" data in the row to display one thing and if there "is not" data in the row to display another. Here is that piece of code. Code: [Select] if ($row['host'] == "") { echo $row['show_name']" with " $row['host']; } else { echo $row['show_name']; } I keep getting this error "Parse error: syntax error, unexpected T_CONSTANT_ENCAPSED_STRING, expecting ',' or ';' in /home/ansonb/lwr_config.php on line 46" What am I doing wrong ? I'm sure it is something simple, I just can't figure it out. Trying to code a first fit algorithm (explanation here). It doesn't output everything correctly at the end. There probably is an easier way to achieve this, but I need to do it using do and while statements. From the output I only get the first array, none of the others display. function first_fit(&$arr, &$max) { $bin = array(array(), array(), array(), array(), array(), array(), array(), array(), array() ,array() ); $i=0; $j=0; do { do { if (array_sum($bin[$j])+$arr[$i] <= $max) {$bin[$j][] = $arr[$i]; $k=1; } $j++; } while($k = 0); $i++; $j=0; } while($i<=9); outputarray($bin[0]); outputarray($bin[1]); outputarray($bin[2]); outputarray($bin[3]); outputarray($bin[4]); outputarray($bin[5]); outputarray($bin[6]); outputarray($bin[7]); outputarray($bin[8]); outputarray($bin[9]); } The function outputarray() just prints the array in a nicely formatted way. Any help would be much appreciated. hi all I am trying to process a result from a query but cant use a echo in a while statement Here is the code I am trying to get the result from each row in the table and then display them I get the error Parse error: syntax error, unexpected T_ECHO in C:\wamp\www\blackrain\Resources\read.php on line 13 <?php $link = mysql_connect('localhost', 'root', ''); if (!$link) { die('Could not connect: ' . mysql_error()); } mysql_select_db("php"); $rs = "SELECT `id`, `title`, `author`, `date`, `imageUrl`, `text` FROM `items` "; $rs_return = mysql_query($rs) or trigger_error($rs . ' has encountered an error: <br />'. mysql_error()); while($obj = mysql_fetch_array($rs_return, MYSQL_ASSOC)) {" <h1>"+ echo $obj['title']; +"</h1> <div id="+"wrapper"+"> <div id="text"> <h4>Posted:"+ echo $obj['author'];+"By:"+ echo $obj['date'] +"</h4> <p class="p3"> <span> <img class="+"alignright"+"src="+ echo $obj['imageUrl'];+"> </img>"+ echo $obj['text'];+"</span></p>" } ?> this if statement isn't working.. $newid is an integer, which is the name of a variable of an array e.g. $53. The below code should check if the array exist. it should echo yes.. instead it is blank. Code: [Select] if(isset(${"$newid"})) { echo "yes"; //code } I've never really had this problem before. I'm trying to make it so only moderators can edit titles. This is currently my if statement: elseif($type == 2 && acc_status(getUsername()) > 1 && !isset($_POST['title']) || !isset($_POST['content'])) (I can assure you getUsername() works) Interpretation: If the editing type is 2 (thread/topic) and the user's account status is higher than 2 (meaning admin & mod) and the title or message/content isn't set, display the content within the if statement block. Why is it users with a user status of 1 can see the information within the block statement, allowing them to see the "edit title" field? I have a query that outputs certain values in a table, but i want to put a conditional if statement. What i have right now is if the status is 1 echo Close and if the status is 0 Open what i want it to do now is not to show the values that have 1 along with all the other database table entries how would i do that Code: [Select] //if statement for status if($rows['status'] == 1) { $status = "Close"; } else $status = "Open"; ?> hi how do u do a if statement so that if <a href="TaxiEntry1.php?Quote_ID='.$qid.'">Accept</a> is clicked then update this field in this table } thanks need help in if statement i will use it wordpress i am semi brand new in php and html i forged it with the help of c and c ++ visual basic LOL and some tutorial but did not work if any one can help making it work <?php $Shreef = <?php echo "get_post_meta($post->ID, 'sub', true);" ?> if ($Shreef=="Subbed"){ <font color="green"> <?php echo "get_post_meta($post->ID, 'sub', true);" ?> </font> } else { if ($Shreef=="Raw") <font color="red"> <?php echo "get_post_meta($post->ID, 'sub', true);" ?> </font> else { <?php echo "get_post_meta($post->ID, 'sub', true);" ?>} } ?> |
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