Display array values in a table

Question

I have the following code, which currently displays the array values in resArr and resArr2 separately. I would like to display this data in a table, showing the value of array resArr in column 1 and the value of resArr2 in column 2. I have tried experimenting with tr and td however for some reason this has no effect on the output.

Could anyone provide me with any possible solution please?

$resArr = json_decode($res, 1);
$resArr2 = json_decode($res2, 1);
foreach ($resArr as $key => $value)
{
    echo "<tr>";
   if(!is_array($value))     
   {
       if($key == 'attire' or $key == 'category' or $key == 'location' or $key == 'name' or $key == 'website' or $key == 'checkins' or $key == 'likes')
        {
            echo "<td>";
            echo  $key . ':' . $value;
            echo "<br />\n";
            echo "</td>";
        }
   }
   echo "</tr>";
}

foreach ($resArr2 as $key => $value)
 {
   //to eliminate array to string conversion error 
   if(!is_array($value))     
   {
       if($key == 'attire' or $key == 'category' or $key == 'location' or $key == 'name' or $key == 'website' or $key == 'checkins' or $key == 'likes')
       {
           echo  $key . ':' . $value;
           echo "<br />\n";
       }
   }
}

You have to iterate through both arrays at the same time: One row(<tr>) will be created at each iteration and you'll add a new column in your loop when going from Arr1 to Arr2 — Answers_Seeker
– Answers_Seeker, Commented May 13, 2015 at 9:46
can you show your expected output screenshot or a glimpse of result hers? — Death-is-the-real-truth
– Death-is-the-real-truth, Commented May 13, 2015 at 10:10

Kuguar6 · Accepted Answer · 2015-05-13 11:17:04Z

1

I belive you want something like this:

$resArr = json_decode($res, 1);
$resArr2 = json_decode($res2, 1);
$labels = array('attire','category','location','name','website','checkins','likes');
$count1=0;

echo '<table border="1">';
echo '<tr><th>key</th><th>resArrLabel</th><th>resArr2Label</th></tr>';
foreach ($labels as $label)
{
    echo '<tr>';
        echo '<th>'.$label.'</th>';
        echo '<td>'.(array_key_exists($label, $resArr) ? $resArr[$label] : '').'</td>';
        echo '<td>'.(array_key_exists($label, $resArr2) ? $resArr2[$label] : '').'</td>';
    echo '</tr>';
}
echo '</table>';

It was much faster becouse use only one loop and add th label to each column and row. Try to add more css style.

edited May 13, 2015 at 11:17

answered May 13, 2015 at 11:07

Kuguar6

261 silver badge5 bronze badges

Sign up to request clarification or add additional context in comments.

9 Comments

user2675041 Over a year ago

The table will still not appear using this method.

Kuguar6 Over a year ago

so try: $resArr = json_decode($res, 1); $resArr2 = json_decode($res2, 1); var_dump($retArr); var_dump($retArr2); die(); and copy result into comment.

user2675041 Over a year ago

array(23) { ["id"]=> string(11) "10936503735" ["about"]=> string(73) "Welcome to asos.do/p3RFnb, where you’ll Discover Fashion Online."["category"]=> string(8) "Clothing" ["checkins"]=> int(2893) ["id"]=> string(17) "10152708202338736" } ["name"]=> string(4) "ASOS" ["parking"]=> array(3) { ["website"]=> string(82) "asos.do/p3RFnb marketplace.asos.com fashionfinder.asos.com" } array(1) { ["error"]=> array(3) { ["message"]=> string(50) "You must use https:// when passing an access token" ["type"]=> string(14) "OAuthException" ["code"]=> int(1) } }

user2675041 Over a year ago

This is only part of the array, had to delete some of it as it did not fit,however it should not make a difference. For some reason the output of resArr2 was not displayed. And I notice that I have an error in the end, however this was not effecting my output when I was not trying to display values in table

Kuguar6 Over a year ago

You must change something else in your code. Try to resolve error. You get data from url without using https protocol. Add https:// at the begin of your url. Output of resArr2 was not displayed becouse there is nothing to display.

|

Shoaib Rehan · Accepted Answer · 2015-05-13 10:15:26Z

1

You have to iterate both arrays at the same time. Try this

$resArr = json_decode($res, 1);
$resArr2 = json_decode($res2, 1);
$count1=0;

echo "<table border='1'>";
foreach ($resArr as $key => $value)
{
	echo "<tr>";
   if(!is_array($value))     
   {
	   if($key == 'attire' or $key == 'category' or $key == 'location' or $key == 'name' or $key == 'website' or $key == 'checkins' or $key == 'likes')
		{
			echo "<td>";
			echo  $key . ':' . $value;
			echo "<br />\n";
			echo "</td>";
		}
   }
   
   $count2=0;
   foreach ($resArr2 as $key => $value)
   {
	   if($count1==$count2){
		 //to eliminate array to string conversion error 
		 if(!is_array($value))     
		 {
			 if($key == 'attire' or $key == 'category' or $key == 'location' or $key == 'name' or $key == 'website' or $key == 'checkins' or $key == 'likes')
			 {
				 echo "<td>";
				 echo  $key . ':' . $value;
				 echo "<br />\n";
				 echo "</td>";
			 }
		 }
	   }
	   $count2++;
  }
   echo "</tr>";
   $count1++;
}
echo "</table>";

edited May 13, 2015 at 10:15

answered May 13, 2015 at 10:05

Shoaib Rehan

5246 silver badges16 bronze badges

5 Comments

user2675041 Over a year ago

The arrays are being iterated at the same time, however the table is not being displayed. The values are just being displayed according to the key

Death-is-the-real-truth Over a year ago

can you show your expected output screenshot or a glimpse of result hers?

user2675041 Over a year ago

I would like to have a table with rows and columns showing the key in the first column, the value of the first array in the second column, and the value of the second array in the third column. However no type of table is being displayed, only plain text

Shoaib Rehan Over a year ago

just add the table tag before and after the parent foreach loop. hope you will see the expected results.

user2675041 Over a year ago

I added this tag echo "<table>" before and after the foreach loop, however a table is still not being displayed.

Shoaib Rehan · Accepted Answer · 2015-05-13 10:54:30Z

1

Look at the results displayed for me. I am using sample array.

$resArr = array('attire'=>'1','category'=>'2','location'=>'3');
$resArr2 = array('attire'=>'a','category'=>'b','location'=>'c');

echo "<table border='1'>";

$count1=0;
foreach ($resArr as $key => $value)
{
	echo "<tr>";
   if(!is_array($value))     
   {
	   if($key == 'attire' or $key == 'category' or $key == 'location' or $key == 'name' or $key == 'website' or $key == 'checkins' or $key == 'likes')
		{
			echo "<td>";
			echo  $key . ':' . $value;
			echo "<br />\n";
			echo "</td>";
		}
   }
   
   $count2=0;
   foreach ($resArr2 as $key => $value)
   {
	   if($count1==$count2){
		 //to eliminate array to string conversion error 
		 if(!is_array($value))     
		 {
			 if($key == 'attire' or $key == 'category' or $key == 'location' or $key == 'name' or $key == 'website' or $key == 'checkins' or $key == 'likes')
			 {
				 echo "<td>";
				 echo  $key . ':' . $value;
				 echo "<br />\n";
				 echo "</td>";
			 }
		 }
	   }
	   $count2++;
  }
   echo "</tr>";
   $count1++;
}
echo "</table>";

Here is the result generated by the above code

answered May 13, 2015 at 10:54

Shoaib Rehan

5246 silver badges16 bronze badges

4 Comments

user2675041 Over a year ago

This is exactly how I would like my code to be shown! I have copied the entire code, however still no table is being shown. Would that mean that the problem is in the array?

Shoaib Rehan Over a year ago

Yah, If this code is not working than make sure your array is correct by print_r($resArr); and print_r($resArr); and comment all the code below.

user2675041 Over a year ago

array(23) { ["id"]=> string(11) "10936503735" ["about"]=> string(73) "Welcome to asos.do/p3RFnb, where you’ll Discover Fashion Online."["category"]=> string(8) "Clothing" ["checkins"]=> int(2893) ["id"]=> string(17) "10152708202338736" } ["name"]=> string(4) "ASOS" ["parking"]=> array(3) { ["website"]=> string(82) "asos.do/p3RFnb marketplace.asos.com fashionfinder.asos.com/"; } array(1) { ["error"]=> array(3) { ["message"]=> string(50) "You must use https:// when passing an access token" ["type"]=> string(14) "OAuthException" ["code"]=> int(1) } }

Shoaib Rehan Over a year ago

do not dump just $resArr = json_decode($res, 1); print_r($resArr); $resArr2 = json_decode($res2, 1); print_r($resArr); die();

Collectives™ on Stack Overflow

Display array values in a table

3 Answers 3

9 Comments

5 Comments

4 Comments

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

3 Answers 3

9 Comments

5 Comments

4 Comments

Your Answer

Sign up or log in

Post as a guest

Related