passing array of string from php to html

Question

I want to tranform a php array of string into html. My php and html code are in the same page.

I have $myvar that hold my array of string. I pass $myvar with POST and insert it to $ba.

My code needs to print on html page 3 line (in while loop). But when I pass the $be, it writes me error message: "Notice: Undefined index: myvar" (in php code)

What do I need to repair so that my code prints to my screen all the 3 lines that I get from php?

my code:(php)

foreach ($docres as $key=>$filename) {
    $counter = 0; 
    $file = $filename +1;
    $handle = fopen($dir."/".$file.'.txt',"r");
    if($handle)
    {
        while($counter < 3)
        {
            $myvar[]=fgets($handle);
            $counter++;
        }
    }
}

$ba = implode("
", $myvar);

my html code:

<form action="" method="POST">
    <center> 
        <h1> My Search Engine </h1>
        <input type = 'text' size='90'  value='' name = 'search' > <br> 
        <input type = 'submit' name = 'submit' value = 'Search source code'>         
    </center> 
</form >
<p> <?php echo $ba ?> </p>

JoelBonetR · Accepted Answer · 2016-09-27 09:44:11Z

1

Simply echo the mysql query on a function and call it on HTML as follows: UPDATE: 3rd column will be an image wich route are stored on database, and 4th col will be an image eich only the name was stored on database (because we know the full route) as example:

 <?php
    function printOnHtml(){
    include ("connection.php");
    $sql = "SELECT * FROM foo;"
    if ($result = connection()->query($sql)){
        $rs = $result->fetch_array(MYSQLI_NUM);
        while ($rs[0] != ''){
           echo "first column: ".$rs[0]." second column: ".$rs[1]." image with full route on database: <img src='".$rs[2]."' alt=''> <br> if only the img name is stored cuz we know the route: <img src='img_route/".$rs[3]."' alt=''>";
           $rs = $result->fetch_array(MYSQLI_NUM);
        }
    }
    }

Then on HTML

<html>
blablabla
<body>
blablabla
<?php
    printOnHtml();
?>
blablabla
</body>
</html>

Note that it have to be a .php file to call the php function (for example index.php)

I paste the connection php script i use in order if you need it:

<?php
function connection(){
if($mysqli = new MySQLi("localhost","user","password","database_name")) echo "OK"; else echo "KO";
mysqli_set_charset($mysqli, "utf8");
return $mysqli;
}
?>

i did it with mysqli fetch array, but you can do the same using fetch assoc if you want.

UPDATE2: If you stubborness makes you follow using a txt to store data (wich, if increase will fail when you get a some thousands line txt), modify this on your code:

$myvar='';
foreach ($docres as $key=>$filename) {
    $counter = 0; 
    $file = $filename +1;
    $handle = fopen($dir."/".$file.'.txt',"r");
    if($handle)
    {
        while($counter < 3)
        {
          if(isset($myvar)){
            $myvar=fgets($handle);
         }
            $counter++;
        }
    }
}

and i'm supposing that you declared $dir, $file and other vars properly. You NEVER have to use vars without declaring it (as NULL at least). You only can do this if you ensured 100% that this var will get a value at this point.

edited Sep 27, 2016 at 9:44

answered Sep 27, 2016 at 9:01

JoelBonetR

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6 Comments

JoelBonetR Over a year ago

oh and... undefined index means just this, you didn't defined the index before using so, if null or wrong, you get this error. Must define it at the top as $myVar = ""; Then you use it after when you need it.

adi Over a year ago

thanks for your answer! my function is not a sql - the files are saved in my storage folder becuse I'm trying to write an inverted index search programe, so in my sql I have word that point to a file. So I got the file I need to print out, and I want to print the first three rows.. how do I do that?

JoelBonetR Over a year ago

to print a file which route is stored in database? let me edit the example and see by yourself

JoelBonetR Over a year ago

Oh and.. you don't need javascripts to reach this job, it's lighting faster using php, specially when dealing with images. =)

adi Over a year ago

lool I did update to the code but its still not working

|

Mr Lister · Accepted Answer · 2016-10-05 20:23:05Z

0

You have to convert the array to string in a correct way using implode and <br> as a separator
Then just print it using php tags (as you are using both at the same page ) you can access the variable direct and print it using <?= $ba ?> or <?php echo $ba ; ?>
Code will be :

<?php
foreach ($docres as $key=>$filename) {
$counter = 0; 
    $file = $filename +1;
    $handle = fopen($dir."/".$file.'.txt',"r");
    if($handle)
    {
        while($counter < 3)
        {
            $myvar[]=fgets($handle);
            $counter++;
        }
    }
}
$ba = implode("<br>", $myvar);
?>


<form action="" method="POST">
    <center> 
        <h1> My Search Engine </h1>
        <input type = 'text' size='90'  value='' name = 'search' > <br> 
        <input type = 'submit' name = 'submit' value = 'Search source code'>         
      </center> 
</form >
<p id="deatiles"> <?= $ba ?> </p>

edited Oct 5, 2016 at 20:23

Mr Lister

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answered Sep 27, 2016 at 9:10

Amr Magdy

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7 Comments

Amr Magdy Over a year ago

@adi are you sure there is a data in $myvar ? please try print_r($myvar); and check if there is a data

Amr Magdy Over a year ago

and please make sure to remove the js code as it is not correct and the $ba at it will be undefined

adi Over a year ago

yes I check it and there is a data. $myvar hold array of string like this: "Array ( [0] => All Of Me ) Array ( [0] => All Of Me [1] => JOHN LEGEND ) "

adi Over a year ago

now its print to my sreen :" <= $ba ?>" and not my data

Amr Magdy Over a year ago

yes sorry i edited the answer from 5 min , it is <?= $ba ?> not <= $ba ?> or you can use <?php echo $ba ?>

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