Need help displaying data from a mysql query

@wevieNov 29.2013

If I search a database and it returns multiple results, what should I use to grab each result so that I can display each one in the browser. Below is the search and I thought I would use the num_rows variable to how a message if no results are found.

[code=php] $result = mysql_query(“SELECT isbn, bookTitle, bookAuthor, thumbnail FROM BooksForSell WHERE ISBN='”.$booksearch.”‘”) or die(mysql_error()); $num_rows = mysql_num_rows($result); [/code]

While the below is not quite formatted the way it will end up looking, I basically want to have a for loop go through the results and show each one in HTML similar to below. I assume that maybe an array would be best for this but I am not sure what the function is to get the results into an array.

[code=php] echo ‘ Title: ‘ .$title. ‘ Author: ‘ . $author . ‘ ISBN: ‘ . $isbn . ‘ Price: ‘ .$price’ Condition: ‘. $condition . ‘ <img src=”.$thumbnail.”>’;

[/code]

to post a comment

PHP

6 Comments(s) _↴

@wevieauthorNov 30.2013 — #So here is where I am not. I found most of the following code at php net but I am still not getting anywhere. Am I even close to doing what I need to do? If I am being honest, I am not really familiar with objects in PHP so that is not helping matters in this case.

[code=php]
 if(mysqli_connect_errno()){
 
   echo "Failed to connect to MySQL: " . mysqli_connect_error();
   }
   $query = "SELECT isbn, bookTitle, bookAuthor, thumbnail FROM BooksforSell WHERE isbn='".$bookSearch."'";
   $result = $mysqli->query($query);
   while($newArray = $result->fetch_array()){
   $isbn = $newArray['isbn'];
   $bookTitle = $newArray['bookTitle'];
   $bookAuthor = $newArray['bookAuthor'];
   $thumbnail = $newArray['thumbnail'];
 
   echo $isbn;
   echo $bookTitle;
   echo $bookAuthor;
   echo $thumbnail;
   }
 [/code]

@wevieauthorNov 30.2013 — #And this should have been the first line in the code only with my database info

[code=php]
 $mysqli = new mysqli("localhost","user,"password","database");
 [/code]

@firesnakerNov 30.2013 — #You have to create a storage array.

<i>
 </i>$aStorage = new Array();
 while($newArray = $result-&gt;fetch_array()){
 $aStorage[] = array(
 'isbn' =&gt; $newArray['isbn'];
 'bookTitle' =&gt; $newArray['bookTitle'];
 'bookAuthor' =&gt; $newArray['bookAuthor'];
 'thumbnail' =&gt; $newArray['thumbnail'];
 );
 }

Then, somewhere else, you create a loop calling the storage array.

<i>
 </i>for ($i = 0; $i &lt; count($aStorage); $i++)
 {
 echo $aStorage[$i]['isbn'];
 echo $aStorage[$i]['bookTitle'];
 echo $aStorage[$i]['bookAuthor'];
 echo $aStorage[$i]['thumbnail'];
 }

If I may suggest you read up the manual section on Array to get a better understanding of the code.

http://us1.php.net/manual/en/language.types.array.php

In short, an array is a pair of key and value. You call the key to retrieve the value.

Now, to store the data you need a multidimensional array. To be specific, a two-dimensional array.

This should be enough to get you started.

Three-dimensional array and beyond are possible, but rarely used in my experience.

Good luck!

@alinarcosNov 30.2013 — #<?php$username="username";

$password="password";$database="your_database";

$field1-name=$_POST['Value1'];

$field2-name=$_POST['Value2'];

$field3-name=$_POST['Value3'];

$field4-name=$_POST['Value4'];

$field5-name=$_POST['Value5'];

mysql_connect(localhost,$username,$password);

@mysql_select_db($database) or die( "Unable to select database");

$query = "INSERT INTO tablename VALUES('','$field1-name','$field2-name',

'$field3-name','$field4-name','$field5-name')";mysql_query($query);mysql_close();?>

@wevieauthorNov 30.2013 — #I tried the code that firesnaker and it did not work. new Array(); seems to be incorrect syntax because that gave me an error. After looking around on php.net, I changed it to just array(), but I was still not getting any data into the array I created.

I have changed my code a bit since the object oriented style is not really making sense to me anyway but I still do not understand how to get the data. I think my problem is that I do not understand how it is returned.

[code=php]
  //creates a query
 $query = "SELECT isbn, bookTitle, bookAuthor, bookCond, thumbnail, price FROM BooksforSell WHERE isbn='".$bookSearch."'";
 //executes a query and returns results
 $result = mysqli_query($dbConnect, $query);
 //assigns the number of rows returned from the query
 $numRows = $result->num_rows;
 $row = mysqli_fetch_array($result, MYSQLI_NUM);
 [/code]

The query I am running in the test returns three rows. I have been looking at php.net and on this if I echo $row[0], $row[1], ......, that will output the first row of data. What I am not understanding is how are the rest of the rows returned? Maybe if I understood that, I could get it.

@wevieauthorDec 01.2013 — #Just thought I would update. I have it working, finally but I did run into one really weird thing that I never did figure out. Below is the code to make the array.

[code=php]
 $query = "SELECT isbn, bookTitle, bookAuthor, bookCond, thumbnail, price FROM BooksforSell WHERE isbn='".$bookSearch."'";
 //executes a query and returns results
 $result = mysqli_query($dbConnect, $query);
 //assigns the number of rows returned from the query
 $numRows = $result->num_rows;
 
   $i = 0; //create index and set to 0
   $bookInfo = array(); //create a new empty array
   //iterate through and fill array with data
   while($row = mysqli_fetch_array($result, MYSQLI_NUM)){
 
   $bookInfo['isbn'][$i] = $row[0];
   $bookInfo['title'][$i] = $row[1];
   $bookInfo['author'][$i] = $row[2];
   $bookInfo['condition'][$i] =$row[3];
   $bookInfo['thumbnail'][$i] = $row[4];
   $bookInfo['price'][$i] = $row[5];
 
   $i++;
 
   }//end while loop 
 [/code]

The weird thing is when I ran the for loop, the output was always twice the index of the array. I have been testing with a query that returns three results so when the for loop ran, it would run six times with the following code.

[code=php]for($i=0; $i < count($bookInfo); $i++)[/code]

I though maybe is was because I used $i in the while loop but it should be local. Either way, I changed the $i to $j and it still did the same thing. The only way I found to get it to work is by using the row count which I set earlier like so.

[code=php]for($i=0; $i < $numRows; $i++)[/code]

Any ideas as to why this was happening?

Also in #PHP _↴

odd result with enbolden after concatenate Locale and Charset fsockopen multithreading modus

Success!

Help @wevie spread the word by sharing this article on Twitter...

Tweet This

Need help displaying data from a mysql query

6 Comments(s) _↴

Also in #PHP _↴

Success!

Social

Version

Need help displaying data from a mysql query

6 Comments(s) ↴

Also in #PHP ↴

Success!

The web is an endless sea of information. Don't miss the boat... Subscribe!

Social

Version

6 Comments(s) _↴

Also in #PHP _↴