PHP - Href Echo Syntax Is Wrong

Code: [Select]
$query = mysql_query("SELECT name FROM business WHERE type ='Restaurant' ORDER BY name");
echo mysql_error();
while($ntx=mysql_fetch_row($query)) $nt[] = $ntx[0];
$i = -1;
foreach($nt as $value) { $i++;
echo $nt[$i]."<br/>";}

Now im trying to swap the last line;
Code: [Select]
echo $nt[$i]."<br/>";}
for somthing more like
Code: [Select]
echo "<a href="$nt[$i].html">$nt[$i]</a>";}
My syntax seems to be shot to pieces, and i recieve errors.  any ideas welcome?

The error i receive is "Parse error: syntax error, unexpected T_VARIABLE, expecting ',' or ';' "

Many thanks.

ok im building facebook app and im just back to php after months not touched it.

this is the code:
$image = $facebook->api_client->photos_get(null,null, echo $uid);
print_r ($image);
echo "<BR>";
echo $image[0]["src"];

the problem is that echo $uid
i tried every variation and it does not work!

i maybe forget something that i should have done?

the $uid containing user id numbers so what i did wrong and if im not wrong i wanted the numbers($uid) Wrapped in quotation marks

so i tried: echo ." $uid ".
and alot of other

please help tnx

Code: [Select]
$height = trim(htmlentities(mysqli_real_escape_string($db,$_POST['height'],ENT_QUOTES)));
Here's the error message I'm getting:

Warning: mysqli_real_escape_string() expects exactly 2 parameters, 3 given

I just added in the htmlentities and ENT_QUOTES to the line but not sure how to format it. Any ideas?

I don't know why but I echoed this and then got the echoed query and put it in the sql of my mysql database and it says
that the syntax isn't right but I looked and I don't see a problem but I am probably over looking something.

$query = "INSERT INTO `characters` 
                (charactername, shortname, status_id, style_id, division_id, alignment_id, sortorder, creator_id, datecreated, enabled) 
            VALUES 
                (".$charactername.",".$shortname.",".$status.",".$style.",".$division.",".$alignment.",".$sortorder.", 1, NOW(), 0)";

Hi all,

As always thanks in advance. My php skills are still at a beginner level but I'm trying to get there!

What I'm trying to do:
Send data to the Mailchimp api (specifically email, name, birthday)

What it currently does do:
Sends and stores email and name

What I cant get it to do:
Send and store the birthday from $birthday string.

First I load the form data:

//Load form data $day = $_POST['Day']; $month = $_POST['Month']; $year = $_POST['Year']; //Create Birthday in mm/dd format for mailchimp $birthday = $month."/".$day;()

Then I pass it later on to mailchimp:

$mergeVars = array('FNAME' => $parsedName['first'], 'LNAME' => $parsedName['last'], 'BDAY'=>"$birthday");()

EVERYTHING in my script works perfectly except for the birthday bit.

So I tried this instead which DOES work, but doesn't use the form data defeating the whole point!

$mergeVars = array('FNAME' => $parsedName['first'], 'LNAME' => $parsedName['last'], 'BDAY'=>"04/04");()

I thought it may be the $birthday string wasn't in the mm/dd setup. But after echo'ing it everything is fine.

I get the feeling I'm doing something stupid here. And I think it's syntax based around the 'BDAY'=>"$birthday"() part.

Any help will be greatly appreciated! This is driving me INSANE.

Thanks,

So, let me explain:
Code: [Select]
<?php
class MySqlDatabase {
private $connection;

function __construct() {
$this->database_connect();
}

public function database_connect() {
$this->connection = mysqli_connect(DB_SERVER, DB_USERNAME, DB_PASSWORD, DB_NAME); 

if (!$this->connection) {
die("Database connection failed: " . mysqli_error());
}

}

public function database_query($sql) {
$query = mysqli_real_escape_string($sql, $this->connection);
$result = mysqli_query($query, $this->connection);
if(!$result) {
die("Database query failed: " . mysqli_error());
}
return $result;
}



public function database_close() {
if (isset($this->connection)) {
mysqli_close($this->connection);
unset($this->connection);
}

}
}
?>
Take a look at the database_query() method.

When I insert this into it from my index.php to test if it works:
Code: [Select]
<?php
$database = new MySqlDatabase();

$sql = "INSERT INTO `website`.`users` (`id`, `username`, `first_name`, `last_name`, `password`, `email`, `secret_question`, `secret_answer`, `create_time`) VALUES (NULL, 'joe', 'joe', 'doe', 'password123', 'myemail@email.com', 'Who am I?', 'myself', '2010-10-16 13:37:59');";

$database->database_query($sql);
?>Obviously the date and the password needs some working on (hashing and entering current time), but I get this error:

Warning: mysqli_real_escape_string() expects parameter 1 to be mysqli, string given in ...includes\classes.php on line 32

Warning: mysqli_query() expects parameter 1 to be mysqli, null given in ..includes\classes.php on line 33

Warning: mysqli_error() expects exactly 1 parameter, 0 given in ..includes\classes.php on line 35
Database query failed:

I I'm giving it a mysql syntax, but it doesn't work, also I get no error message.

Hi,
I am no PHP expert. I can poke about, do a little cut and paste.
Dreamweaver is telling me I have a syntax issue with my if statement. I have tried a few things without success.
Dreamweaver says the error is on the else line, that may not be true. please show me what I am doing wrong.
Thanks Mark
Code: [Select]
<div class="entry-content tec-event-entry">
<?php /** Get the "cft_tinymce_1" custom field */
 $block_1 = get_custom_field('cft_tinymce_1');
if( $block_1 ) { ?>
<?php echo $block_1; ?>
<?php } ?>
<?php else { ?>
<?php the_content(); ?>
<?php } ?>
</div> <!-- End tec-event-entry -->

Hi guys; 

I've managed to find my way into something of a maze. 

<td> <a href="'Details/21/index.php'"?id= . $id .'>" . $row['id'] . "</a> </td> 

I'm looking at this line in the code. I realise it's currently in the wrong syntax but I'm just trying multiple different variations.

At the moment I'm actually just trying to make it go to a static link but my real goal is to - get it to pick up the id number, of the id row I click, and concatenate that with the rest of my address string. Then use that as the href. 

Something like this: -

"'Details/' + $id + '/index.php'"

It's really confusing me though inside this loop and for some reason the id number is being picked up as an int, by the looks of things. It's getting above my level of understanding. 

Any chance one of you masters would through a n00b a lifeline? 

 

This is my code below 

 

<!DOCTYPE html>
<html>
<head>
	<title>LifeSaver DB</title>
	<h1> LifeSaver Database </h1>
</head>
<body>
	<table>
		<tr>
			<th>Id</th>
			<th>Location</th>					
			<th>Initials</th>
			<th>TimeStamp</th>
			<th>Notes</th>					
		</tr>
		<?php
		$conn = mysqli_connect("localhost", "meh", "pas", "DB"); 
		// Check connection
		if ($conn->connect_error) {
			die("Connection failed: " . $conn->connect_error);
			}
			
		$sql = "SELECT * FROM LifeSaver1 ORDER BY id DESC";
		$result = $conn->query($sql);
		
		if ($result->num_rows > 0) {
			// output data of each row
			while($row = $result->fetch_assoc()) {
				
				//for href row
				$id = $row['id'];
				$Footage = ['Footage'];
				
				echo 
				"<tr>
					<td> <a href="'Details/21/index.php'"?id= . $id .'>" . $row['id'] . "</a> </td>                
					<td>" . $row["Location"] . "</td>		
					<td>" . $row["Initials"]. "</td>
					<td>" . $row["TimeStamp"]. "</td>	
					<td>" . $row["Notes"] . "</td>
							
				</tr>";}
				
				//show table
				echo "</table>";
				} else { echo "0 results"; }
			$conn->close();

	?>
	</table>
</body>
<style>
	
table, td, th {
  border: 1px solid black;
  margin: auto;
}

table {
  border-collapse: collapse;
color: #000;   <!--font colour -->
font-family: monospace;
font-size: 18px;
text-align: center;}

th {
background-color: #337AFF;
color: white;
font-weight: bold; }

tr:nth-child(odd) {background-color: #add8e6}


</style>
</html>

 

Edited February 8, 2020 by JonnyDriller

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;

?>

I have been looking at this code most of the morning and do not have a clue what is wrong with the code. I am hoping its not a stupid mistake, can someone please help me out? thank you   

 
<title>Inputing Travel Detials</title>
 
<header>
  <h1 align="center"> Adding Travel Detials </h1>
  
 
  <body>
    <p>  <center><img  src="cyberwarfareimage1.png" alt="Squadron logo" style="width:200px;height:200px" style="middle"></center>
        <table border="1">
          <tr>
            <td><a href="index.php"> Home Page </a></td>
            <td><a href="administratorhomepage.html">Administrator Home Page </a></td>
            <td><a href="viewhomepage.html">View Home Page </a></td>
            <td><a href="Inputhomepage.html">Input Home Page </a></td>
            <td><a href="traveldetials.html">Enter More Travel Detials </a></td>
        </table>
    </p>
 
 
<?php
 
include "connection.php";
 
$Applicant_ID = $_POST["Applicant_ID"];
$Method_Of_Travel = $_POST["Method_Of_Travel"];
$Cost = $_POST["Cost"];
$ETA = $_POST["ETA"];
$Main_Gate_Advised = $_POST["Main_Gate_Advised"];
 
$query = ("UPDATE `int_board_applicant` SET `Method_Of_Travel`=`$Method_Of_Travel', `Cost`=`$Cost', `ETA`='$ETA', `Main_Gate_Advised`='$Main_Gate_Advised' WHERE `Applicant_ID`='$Applicant_ID'");
   
 
$result = mysqli_query($dbhandle, $query) 
    or die(mysqli_error($dbhandle));
 
if($result){ echo "Success!"; }
 
  else{ echo "Error."; }
 
 
// successfully insert data into database, displays message "Successful".
 if($query){
  echo "Successful";
}
 
  else {
  echo "Data not Submitted";
}
 
 
//closing the connection
   mysqli_close($dbhandle)
   
 
 
 
 
?>

Ok this is puzzleing. I am geting "Could not delete data: You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near '1' at line 1". but its is deleting the entry that needs to be removed. The "1" is the entry. Just not sure what is causing the error. I do have another delete php but I have put that on the back burning for the time being.     <?php

         $con = mysqli_connect("localhost","user","password","part_inventory");
    // Check connection
if (mysqli_connect_errno()) {
  printf("Connect failed: %s\n", mysqli_connect_error());
    exit();
} else {
    
    $result = mysqli_query($con, "SELECT * FROM amp20 ");
    
    $amp20ptid = $_POST['amp20ptid'];                // escape variables for security
    $amp20ptid = mysqli_real_escape_string($con, $_POST['amp20ptid']);
    
    mysqli_query($con, "DELETE FROM amp20 WHERE amp20ptid = '$amp20ptid'");
      
    
if (!mysqli_query($con, $amp20ptid));

{
 die('Could not delete data: ' . mysqli_error($con));    
}
 echo "Part has been deleted to the database!!!\n";
    mysqli_close($con);
}
?>

Hi guys   I have this code below and all works fine when submitting this online application apart from when someone types either ' # & into one of the comment fields in which it throws up the error. Have tried various fixes from across the internet but no joy. Can anyone offer suggestions?   <?php $con = mysql_connect("localhost:3306","root","password"); if (!$con)   {   die('Could not connect: ' . mysql_error());   } mysql_select_db('sfapp', $con); $sql="INSERT INTO 'sfapp' ('surname_add','forename_add','dob_add','hometele_add','mobiletele_add','homeadd_add','siblings_add','schoolname_add','headname_add','schooladd_add','schooltele_add','schoolem_add','alevel_add','personstate_add','nameprovided_add','pe_add','se_add','PredGrade_Art','PredGrade_AScience','PredGrade_BusStudies','PredGrade_Electronics','PredGrade_EnglishLang','PredGrade_EnglishLit','PredGrade_French','PredGrade_German','PredGrade_Geog','PredGrade_Graphics','PredGrade_History','PredGrade_Maths','PredGrade_SepScience','PredGrade_ProductDesign','PredGrade_Spanish','PredGrade_Other','Gender_Male','Gender_Female','Sub_EnglishLit','Sub_Maths','Sub_FurtherMaths','Sub_Biology','Sub_Chemistry','Sub_Physics','Sub_French','Sub_German','Sub_Spanish','Sub_Geography','Sub_History','Sub_RE','Sub_FineArt','Sub_Business','Sub_Computing','Sub_GlobPersp','Sub_DramaAndTheatre','Sub_PE','Sub_Dance','Sub_Politics','Sub_Psychology','Sub_Sociology','readprospect_chk','Sib_Yes','Sib_No','Current_Student_Yes','Current_Student_No','I_Understand_chk','Current_Education_chk','Local_Care_chk','Staff_Cwhls_chk','Sub_Film')   VALUES ('$_POST[surname_add]','$_POST[forename_add]','$_POST[dob_add]','$_POST[hometele_add]','$_POST[mobiletele_add]','$_POST[homeadd_add]','$_POST[siblings_add]','$_POST[schoolname_add]','$_POST[headname_add]','$_POST[schooladd_add]','$_POST[schooltele_add]','$_POST[schoolem_add]','$_POST[alevel_add]','$_POST[personstate_add]','$_POST[nameprovided_add]','$_POST[pe_add]','$_POST[se_add]','$_POST[PredGrade_Art]','$_POST[PredGrade_AScience]','$_POST[PredGrade_BusStudies]','$_POST[PredGrade_Electronics]','$_POST[PredGrade_EnglishLang]','$_POST[PredGrade_EnglishLit]','$_POST[PredGrade_French]','$_POST[PredGrade_German]','$_POST[PredGrade_Geog]','$_POST[PredGrade_Graphics]','$_POST[PredGrade_History]','$_POST[PredGrade_Maths]','$_POST[PredGrade_SepScience]','$_POST[PredGrade_ProductDesign]','$_POST[PredGrade_Spanish]','$_POST[PredGrade_Other]','$_POST[Gender_Male]','$_POST[Gender_Female]','$_POST[Sub_EnglishLit]','$_POST[Sub_Maths]','$_POST[Sub_FurtherMaths]','$_POST[Sub_Biology]','$_POST[Sub_Chemistry]','$_POST[Sub_Physics]','$_POST[Sub_French]','$_POST[Sub_German]','$_POST[Sub_Spanish]','$_POST[Sub_Geography]','$_POST[Sub_History]','$_POST[Sub_RE]','$_POST[Sub_FineArt]','$_POST[Sub_Business]','$_POST[Sub_Computing]','$_POST[Sub_GlobPersp]','$_POST[Sub_DramaAndTheatre]','$_POST[Sub_PE]','$_POST[Sub_Dance]','$_POST[Sub_Politics]','$_POST[Sub_Psychology]','$_POST[Sub_Sociology]','$_POST[readprospect_chk]','$_POST[Sib_Yes]','$_POST[Sib_No]','$_POST[Current_Student_Yes]','$_POST[Current_Student_No]','$_POST[I_Understand_chk]','$_POST[Current_Education_chk]','$_POST[Local_Care_chk]','$_POST[Staff_Cwhls_chk]','$_POST[Sub_Film]')"; if (!mysql_query($sql,$con))   {   die('Error: ' . mysql_error());   }   ?> <?php //if "email" variable is filled out, send email   if (isset($_REQUEST['pe_add']))  {     //Email information     $admin_email = $_REQUEST['pe_add'];   $forename = $_REQUEST['forename_add'];   $email = "autoreply@testing.com";   $subject = "Application";   $desc =   "Dear $forename   Thank you for submitting your online application, we will be in touch shortly. " ;       //send email   mail($admin_email, "$subject", "$desc", "From:" . $email);     //Email response   echo "Thank you for contacting us!";   }     //if "email" variable is not filled out, display the form   else  { ?> If you are seeing this, you need to go back and fill out the Personal Email section! <?php   }   header("location:complete.php"); mysql_close($con) ?>     Thanks in advance.

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']}; 
?>

Hi..

I create mysql syntax for query testing before i input to my php code

here is my mysql code:

Code: [Select]
set @t = 0;
set @rqty=31968;
SELECT LOT_CODE as code,  DATE_ENTRY,   
CASE WHEN @t+OUTPUT_QTY > @rqty
THEN @rqty  -@t 
ELSE OUTPUT_QTY
END as qty,
@t := @t + d.OUTPUT_QTY as cumulative   
FROM dipping d  WHERE SUBSTR(LOT_CODE, 9,4) = 'P28' AND (@t < @rqty);

and i attach the sample output of the above query.

Now that query test is work i  will input that code to my php codes.


     $sql = "SELECT SKUCode, Materials, Comp, Qty 
     FROM bom  WHERE SKUCode = '$SKUCode'";
     $res = mysql_query($sql, $con);

    ($row = mysql_fetch_assoc($res));
     $Materials = $row['Materials'];
     $Qty = $row['Qty'];
     $Comp = $row['Comp']; //P28
//-----Compute Req Qty and Save to table---//     

     $ReqQty = $Qty * $POReq; // 31968


     
$sql = "UPDATE bom SET ReqQty = '$ReqQty' WHERE SKUCode = '$SKUCode' AND Materials = '$Materials'";
$resReqQty = mysql_query($sql, $con);

$t = 0;

$sql = "SELECT LOT_CODE as code,  DATE_ENTRY,   
CASE WHEN $t+OUTPUT_QTY > $ReqQty 
THEN $ReqQty  -$t  
ELSE OUTPUT_QTY 
END as qty,
$t := $t + d.OUTPUT_QTY as cumulative   
FROM dipping d  WHERE SUBSTR(LOT_CODE, 9,4) = '$Comp' AND ($t < $ReqQty)";



when I echo the query:

I got this:

 SELECT LOT_CODE as code, DATE_ENTRY, CASE WHEN 0+OUTPUT_QTY > 31968 THEN 31968 -0 ELSE OUTPUT_QTY END as qty, 0 := 0 + d.OUTPUT_QTY as cumulative FROM dipping d WHERE SUBSTR(LOT_CODE, 9,4) = 'P28' AND (0 < 31968)

then I run it to the sql

and I got an error:

Error Code : 1064
You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near ':= 0 + d.OUTPUT_QTY as cumulative FROM dipping d WHERE SUBSTR(LOT_CODE, 9,4) = '' at line 1
(0 ms taken)


Any help is highly appreciated

Thank you so much

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?