PHP - Getting A Query Field To Display As A Hyperlink

Hi there.  I'm totally new (about a week!) with php and mysql and am encountering a problem that perhaps someone can help me with? I've looked through to see if a similar problem has appeared or been solved, but without success, so apologies if I am repeating something.    In php I am trying to update 7 fields from a form from which a user has edited/modified any of the fields in a chosen record (except id).   Here is the code:   $id=$_GET['id'];

ok, this is clearly 1st grade code for some, but i'm not there yet - I'm querying posts in WordPress, and the post_content will always have an image in the beginning of the post followed by the content. i don't want to get the image, just the content that's after the image, which is wrapped in anchor tags, of course.

Code: [Select]
<a href="http://path/to/image.jpg"><img src="http://path/to/image.jpg" /></a> <p>Post content yadda, yadda, hoowie</p>


obviously a character count won't work, so i need to get anything that follows the first "</a>", say...? is this the best way, or is there an easier way?

thanks for anyone's help.

GN

Hi, I'm trying to set up a page which first queries for mySQL record results matching a country that the user selects. This works fine, but in the event of there being no records for that country, I want it to look at another field, "Region" and pick the records matching that Region instead.

For example, a user searches for "Australia" but there are no matching records. So, I want it to pick all the records for the region of Australasia, and display records for Australia, New Zealand, Papua New Guinea and so on.

I had created the following:


$query = "SELECT * FROM specialists WHERE Country LIKE '$country' ORDER BY SpecialistName"; // specify the table and field names for the SQL query
 //$query .= " limit $s,$limit";
 $numresults=mysql_query($query);
 $numrows=mysql_num_rows($numresults);
 
if ( $numrows == 0 ) { 
echo '<p>We don\'t have any results for specialists in countries near to yours at the moment. Please try <a href="specialists.php" style="text-decoration:underline;">searching a different country</a></p>'; 

} 
// get results
  $result = mysql_query($query) or die("Couldn't execute query");
// display the results returned
while ($row= mysql_fetch_array($result)) {
  $region = $row["Region"];
$count++ ;
}
// next determine if s has been passed to script, if not use 0
  if (empty($s)) {
  $s=0;
  }
echo '<table width="600" class="cardisplay"><tr>';
$count = 1 + $s ;
echo $region;
// Build SQL Query  
$query2 = "SELECT * FROM specialists WHERE Region LIKE '$region' ORDER BY SpecialistName"; // specify the table and field names for the SQL query
 //$query .= " limit $s,$limit";
 $numresults=mysql_query($query2);
 $numrows=mysql_num_rows($numresults);

// get results
  $result = mysql_query($query2) or die("Couldn't execute query");
// display the results returned
while ($row= mysql_fetch_array($result)) {
  $title1 = $row["ID"];
  $specialistname = $row["SpecialistName"];
  $address1 = $row["Address1"];
  $address2 = $row["Address2"];
  $address3 = $row["Address3"];
  $address4 = $row["Address4"];
  $address5 = $row["Address5"];
  $postcode = $row["Postcode"];
  $country = $row["Country"];
  $region = $row["Region"];
  $website = $row["Website"];
  $email = $row["Email"];
  $telephone = $row["Telephone"];
  $businesstype = $row["BusinessType"];

//followed by echoing out the various data etc. etc.



But the problem is that $region is always blank / empty in the cases where $query is empty, so I can't pull the value out and therefore $query2 is also empty...

Any ideas?

I created a drop-down menu using a MySQL statement in php for a form.
My drop down menu works fine, but I want to assign a default value to it (normal text) the value will never change,
thus I do not need to extract the default value from the table, I already know the value.

Here is the basic snippet from the script:

<?php
   include("../includes/xxx.php");
   $cxn = mysqli_connect($host,$user,$password,$dbname);
   $query = "SELECT DISTINCT `plant_id` FROM `plant` ORDER BY `plant_id`";
   $result = mysqli_query($cxn,$query);
      while($row = mysqli_fetch_assoc($result))
      {
      extract($row);
      echo "<option value='$plant_id'>$plant_id</option>\n";
      }
?>

I have a form that allows my client to update some products. Now the products are simple just basic info and 1 picture.

I have set this up so they can edit the products and change the information, having done this many times in the past, but now hit a puzzling block that I am baffled.

The client when editing is presented with the form with the information pulled from the database and the form fields loaded with that data ready to edit.

The image can either be left alone or they can choose to upload a new image. They are shown the image they currently have stored in the database.

The problem I have is EVEN if they decide not to upload an image and change other information, when the submit the form it must be sending a blank value for the image somewhere as it is updating the database and removing the image reference as if it has been removed.

I have an if/else statement based on the form to perform 2 different queries for the update in mysql.

Here is the code for the form update, as you can see the image should not update?? Please help??


 if ($_SERVER['REQUEST_METHOD'] =='POST') {

//This stops SQL Injection in POST vars 
 foreach ($_POST as $key => $value) { 
  $_POST[$key] = mysql_real_escape_string($value); 
 }

// **************************** THIS IS FOR NO NEW IMAGE ********************************
 
 if ($_SERVER['REQUEST_METHOD'] =='POST' && empty($_FILES['product_image']['name'])) {

   # setup SQL statement for no new image
  $SQL = " UPDATE products SET product_title = '{$_POST['product_title']}', product_description = '{$_POST['product_description']}', standard_price = '{$_POST['standard_price']}', deluxe_price = '{$_POST['deluxe_price']}' WHERE product_id = '{$_REQUEST['product_id']}' ";
}

// **************************** THIS IS FOR A NEW IMAGE ********************************

 else {

// Check the image type is a jpeg or gif for the image.
 if (($_FILES['product_image']['type'] != "image/gif") && ($_FILES['product_image']['type'] != "image/jpeg") && ($_FILES['product_image']['type'] != "image/pjpeg")) {
  echo "<FONT FACE=\"Verdana\"><SPAN CLASS=\"content\">You have chosen not to upload a <b>Product Image</b>.<BR></SPAN>" ;
 } elseif ($_FILES['product_image']['size'] > 100000) {
  echo "<FONT FACE=\"Verdana\"><SPAN CLASS=\"content\">The file size is bigger than 300kb.<BR></SPAN>" ;
 } else {
  move_uploaded_file($_FILES['product_image']['tmp_name'], "/httpdocs/product_images/".$_FILES['product_image']['name']) ;
 echo "<FONT FACE=\"Verdana\"><SPAN CLASS=\"content\"><B>Your front image has successfully uploaded.</B><BR></SPAN>" ;
  }
}

  # setup SQL statement for update
  $SQL = " UPDATE products SET product_title = '{$_POST['product_title']}', product_description = '{$_POST['product_description']}', standard_price = '{$_POST['standard_price']}', deluxe_price = '{$_POST['deluxe_price']}', product_image = '{$_FILES['product_image']['name']}' WHERE product_id = '{$_REQUEST['product_id']}' ";
}
  
  #execute SQL statement
  $result = mysql_db_query( *****,"$SQL",$connection );

  # check for error
  if (!$result) { echo("ERROR: " . mysql_error() . "\n$SQL\n");    }

I have a MySQL table with a list of albums and there is a field called "views" with the number of views each album has received. I'm looking to generate an array of all the albums in the table and sort the array by the number of views (descending). I have a list of functions defined in a ContentController.php file. I created a new function called build_albumlist, which I've pasted below. The function "get_ip_log" already exists and works, and I used it as a template to create the "build_albumlist" function:





   public function build_albumlist(){
      return $this->select_raw("SELECT * FROM albums WHERE deleted = '0' ORDER BY views DESC",array(),'all');
   }

public function get_ip_log(){
      return $this->select_raw("SELECT * FROM sessions ORDER BY ID DESC",array(),'all');
   }


When I use the function, I get this warning:


Warning: mysql_real_escape_string() expects parameter 1 to be string, array given inC:\xampp\htdocs\Controllers\DBController.php on line 10

[font=cabin, 'trebuchet ms', helvetica, arial, sans-serif]The "select_raw" function that I used in "build_albumlist" is defined in the DBController.php file, and is defined as below:[/font]
private function clean_array($params){
      $final=array();
      foreach($params as $param){
         $final[]=mysql_real_escape_string($param);
      }
      return $final;
   }
   public function select_raw($query,$params,$type=''){
      $query=str_replace("?","'%s'",$query);
      $final_query= call_user_func_array('sprintf', array_merge((array)$query, $this->clean_array($params)));
      if($type==''){
         $result=mysql_query($final_query) or die(mysql_error());
         return mysql_fetch_assoc($result);
      }
      elseif($type=='all'){
         $result=mysql_query($final_query) or die(mysql_error());
         $final=array();
         while($row=mysql_fetch_assoc($result)){
            $final[]=$row;
         }
         return $final;
      }


Does anyone know why the "build_albumlist" function is generating this warning, while the "get_ip_log" is not? Any help would be great, as I am obviously pretty new to this.

I would like to echo field names in my table on webpage one by one. I know mySQL has "describe" function which will list the complete table, I am looking for a way to display each field name one by one with other stuff in between them like input field or description.

Hi all,

I have the following MySQL insert query:
Code: [Select]
$insert= mysql_query   ("INSERT INTO tablename (column1,`".$EXPfields."`) VALUES ('$something','".$EXPvalues."')");
where $EXPfields is an array of table-field-names and $EXPvalues is an array of table-field-values.

Now I want to write an equivalent query, but using UPDATE instead of INSERT INTO, but I don't want to write out all the field names/values separately, but again want to use $EXPfields and $EXPvalues.

So something like this:
Code: [Select]
$update = mysql_query ("UPDATE tablename
                     SET (column1,`".$EXPfields."`) = ('$something','".$EXPvalues."')
                     WHERE ....
                  ");

Is this possible? If so, what is the proper syntax? Thanks!

I know that there is a way to this with javascript, but I don't know how it would work.  I am much more familiar with php so I figured I would ask here to see if anyone might have a good php solution.

I'm trying to have a form text field display depending on whether or not the user checks a specific box, but I'm not sure where to start.  I've searched around but I didn't find anything that would help.  Basically I want to have a checkbox that asks if the user would like to link their account to an email address.  If the checkbox is selected, a textbox will then appear on the page for them to enter an email address.  Can anyone point me in the right direction on where to start?  Thanks for your help.

Hi,

I'm new to PHP so sorry if it is difficult to understand my question. I've just added a new custom field for my website home page (home.php). I've image link inside that custom field and now I want to display that image at Home (Home.php ).

Please let me know what code/loop should I add in home.php to get that image display with size of 100*50 px .

I hope you understand what I meant to.

Thank You.

hi im trying to display the form entities in the textarea field so they can be sent to the user in an email, but its not working properly it prints out the variable $idName instead of its value?? any help would be greatly appricated ive been stuck on this bit of ages


<?php
function display_output_page()
{
   $idName = trim($_REQUEST['username']);
   $firstName = trim($_REQUEST['firstname']);
   $Surname = trim($_REQUEST['surname']);
   $Address = trim($_REQUEST['address']);
   $Email = trim($_REQUEST['email']);
   $Phone = trim($_REQUEST['phone']);
   $Fitness = isset($_REQUEST['fitness']) ? $_REQUEST['fitness'] : '';
   $goGym = isset($_REQUEST['gogym']) ? $_REQUEST['gogym'] : '';
   $whyGym = isset($_REQUEST['whyGym']) ? $_REQUEST['whyGym']: array();
?>
   <html>
   <head><title>Form Results</title></head>
   <body>
   <h1>Form Results</h1>
   <form action='mailto:$Email?subject=user comments' method='post'enctype='text/plain'>
<textarea name='Topic' rows=15 cols=90>

idName: ". $idName\n;
Firstname: ". $_POST["firstname"]\n;
Surname: ". $_POST["surname"]\n;
Address: ". $_POST["address"]\n;
Email: ". $_POST["email"]\n;
Phone no.: ". $_POST["phone"]\n;
Your fitness level: ". $_POST["fitness"]\n;
How often do you go to the gym: ". $_POST["goGym"]\n;
Why do you go to the Gym: ". $_POST["whyGym"];
</textarea></br>
<input type='submit' value='Send Thru Email'>
</form>
</body>
</html>
<?php
}
?>

Howdy,

I'm trying to display text from a table in a database. It's a list of quotes, so I just need to pull out the quote and the author name. However, the quote and name are not fields in the same record; they are separate records. Example data:

Code: [Select]
quoteid    name        value
1            content You guys are great! Thanks for being awesome.
1            author John Jackson
2            content Gosh you're amazing! Always been so darn helpful!
2            author Peter Davis
So, I just need to rip out the data from 'content' and 'author', and then group them together based on the quoteid. This is my code thus far:

$testimonial_resource = mysql_query("SELECT name, value FROM quotes GROUP BY quoteid ORDER BY author ASC") or die(mysql_error());

while ($testimonial = mysql_fetch_assoc($testimonial_resource)) {

echo '<p>'.$testimonial['value']'.<br /><strong>'.$testimonial['value'].'</strong></p>';

}

Any help would be greatly appreciated for this novice.

Cheers.

Hi guys, I have a problem. I need to create a page that has a web form to upload an image to a mySQL database and place it in a blob field. Then I need to be able to query the database later to display the image on the site. I've looked around but I just haven't found any examples that can help me. Does anyone know of any good example or can anyone please give me an example? I have nothing so far, just the html web form and database.

this is my 3 files which is i am using to show data of customer
but i am not able to see them their logo image please see them & help me guys

i dont know whats wrong in this.

these files i uploaded please find attachment to see them.

thnx in advanced

[attachment deleted by admin]

Alright, I looked though the read me's, went over the FAQ's... Think it is time to post...   So, this lump is what I got from going though google and piecing together the bits that looked good. 

<?php
$db_host = "****";
$db_user = "****";
$db_pwd = "****";
$database = "****";
$table = "****";
$query = "SELECT * FROM {$table}";
$conn = mysqli_connect($db_host, $db_user, $db_pwd, $database);
$result = mysqli_query($conn,$query) or trigger_error($query . ' - has encountered an error at:<br />' . mysqli_error($conn));
$fields = mysqli_fetch_fields($conn);
$field_count = mysqli_num_fields($conn);
echo '<table border="1" style="width:100%">' . "\n" . '<tr>';
$i = 0;
foreach($fields as $field)
{
	if(++$i == 1) {
		echo '<th colspan="' . $field_count . '">' . $field->table . '</th></tr><tr>';
	}
echo "\n" . '<th>' . $field->name . '</th>';
}
echo '</tr>';
while($row = mysqli_fetch_row($result)) {
	echo '<tr><td>' . implode('</td><td>' , $row) . '</td></tr>';
}
echo '</table>';





?>

The goal is to display the table and conditionally format the contents.     Let's forget about the conditional part (thinking if/than but don't know how I'm going to do that yet) and focus on the key problem: displaying the table.   I have tested this code. It works...to a point. it displays the table and its contents but no headers. I know my issue is in line 20 and 22 but I can not for the life of me figure it out. If it is stupidly easy (for you) please remember that I am not a coder. At best I am a scripter when it comes to linux. I am over my head on this stuff.   Thank you all for any help you can give me
Edited by TheAlmightyOS, 18 November 2014 - 03:30 PM.