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Query and display MySQL DB data from HTML

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0

I want to link MySQL DB to the HTML and get the data according to the query below.

SELECT * FROM feature1 WHERE place = 'feature2'

php seems one of the way to do this, since I am one day old with php don't have much knowledge. I wrote the code below, but when run it gives me an error with no solid output. Is the code correct?

I want to query the DB using the above query and display data!!

<?php
$db = mysqli_connect('localhost','user','password','DB_name')
or die('Error connecting to MySQL server.');

> $query = "SELECT * FROM feature1";


mysqli_query($db, $query) or die('Error querying database..');
$result = mysqli_query($db, $query);
$row = mysqli_fetch_array($result);
while ($rows = mysql_fetch_array($query)) { 
 $name = $rows['Name'];
 $address = $rows['Address'];
 $email = $rows['Email']

 echo "$name<br>$address<br>$email<br>$subject<br>$comment<br><br>"; 
?>

javascript php html mysql

share|improve this question

edited 2 days ago

Vince

14.8k32849

asked Apr 1 at 16:20

AngelineAngeline

1

New contributor

Angeline is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  if using php you need to pull the data out a database and turn it into a html friendly format (xml/json) like this > stackoverflow.com/questions/5112282/…
  
  – Mapperz♦
  Apr 1 at 16:38
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Yes will try that thanks.
  
  – Angeline
  yesterday
  

  
  
  
  

add a comment | 

0

I want to link MySQL DB to the HTML and get the data according to the query below.

SELECT * FROM feature1 WHERE place = 'feature2'

php seems one of the way to do this, since I am one day old with php don't have much knowledge. I wrote the code below, but when run it gives me an error with no solid output. Is the code correct?

I want to query the DB using the above query and display data!!

<?php
$db = mysqli_connect('localhost','user','password','DB_name')
or die('Error connecting to MySQL server.');

> $query = "SELECT * FROM feature1";


mysqli_query($db, $query) or die('Error querying database..');
$result = mysqli_query($db, $query);
$row = mysqli_fetch_array($result);
while ($rows = mysql_fetch_array($query)) { 
 $name = $rows['Name'];
 $address = $rows['Address'];
 $email = $rows['Email']

 echo "$name<br>$address<br>$email<br>$subject<br>$comment<br><br>"; 
?>

javascript php html mysql

share|improve this question

edited 2 days ago

Vince

14.8k32849

asked Apr 1 at 16:20

AngelineAngeline

1

New contributor

Angeline is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  if using php you need to pull the data out a database and turn it into a html friendly format (xml/json) like this > stackoverflow.com/questions/5112282/…
  
  – Mapperz♦
  Apr 1 at 16:38
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Yes will try that thanks.
  
  – Angeline
  yesterday
  

  
  
  
  

add a comment | 

I want to link MySQL DB to the HTML and get the data according to the query below.

SELECT * FROM feature1 WHERE place = 'feature2'

php seems one of the way to do this, since I am one day old with php don't have much knowledge. I wrote the code below, but when run it gives me an error with no solid output. Is the code correct?

I want to query the DB using the above query and display data!!

<?php
$db = mysqli_connect('localhost','user','password','DB_name')
or die('Error connecting to MySQL server.');

> $query = "SELECT * FROM feature1";


mysqli_query($db, $query) or die('Error querying database..');
$result = mysqli_query($db, $query);
$row = mysqli_fetch_array($result);
while ($rows = mysql_fetch_array($query)) { 
 $name = $rows['Name'];
 $address = $rows['Address'];
 $email = $rows['Email']

 echo "$name<br>$address<br>$email<br>$subject<br>$comment<br><br>"; 
?>

javascript php html mysql

share|improve this question

edited 2 days ago

Vince

14.8k32849

asked Apr 1 at 16:20

AngelineAngeline

1

New contributor

Angeline is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

SELECT * FROM feature1 WHERE place = 'feature2'

<?php
$db = mysqli_connect('localhost','user','password','DB_name')
or die('Error connecting to MySQL server.');

> $query = "SELECT * FROM feature1";


mysqli_query($db, $query) or die('Error querying database..');
$result = mysqli_query($db, $query);
$row = mysqli_fetch_array($result);
while ($rows = mysql_fetch_array($query)) { 
 $name = $rows['Name'];
 $address = $rows['Address'];
 $email = $rows['Email']

 echo "$name<br>$address<br>$email<br>$subject<br>$comment<br><br>"; 
?>

share|improve this question

edited 2 days ago

Vince

14.8k32849

asked Apr 1 at 16:20

AngelineAngeline

1

New contributor

Angeline is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

share|improve this question

edited 2 days ago

Vince

14.8k32849

asked Apr 1 at 16:20

AngelineAngeline

1

New contributor

Angeline is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

edited 2 days ago

Vince

14.8k32849

edited 2 days ago

Vince

14.8k32849

asked Apr 1 at 16:20

AngelineAngeline

1

New contributor

Angeline is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

asked Apr 1 at 16:20

AngelineAngeline

1