PHP getting result from foreach loop

Question

I am new to PHP I am trying to get the foreach loop looping result. The loop I have

foreach($results as $result){
   $result = $result['names'];
}

Inside the loop are strings (John, Fred, Ann);

I am trying to get this outside the loop, when I try to echo $result I only have John, what should I do to get all three names?

I have tried to create an array like

$resultData= array();
foreach($results as $result){
  $resultData = $result['names'];
}
echo $resultData;

This doesn't work, does anyone have any suggestions?

Updated question: When I try to var_dump($resultData) I am getting

string(7) "John" string(7)"Fred" string(7)"Ann"

you should consult the manual more in this topic, this is just simple array usage — Kevin
– Kevin, Commented Mar 8, 2016 at 3:45
Thank you for all your replies, this loop is an multidimensional array, I am trying to get a single loop result out, so I don't think explode the loop is what I wanted. — user999
– user999, Commented Mar 8, 2016 at 4:03

XsiSecOfficial · Accepted Answer · 2016-03-08 09:26:00Z

2

Here is an example too:

$employeeAges;
$employeeAges["Lisa"] = "28";
$employeeAges["Martin"] = "16";
$employeeAges["Erik"] = "35";
$employeeAges["Victor"] = "46";
$employeeAges["Grace"] = "34";

foreach( $employeeAges as $key => $value){
    echo "Name: $key, Age: $value <br />";
}

Output
Name: Lisa, Age: 28
Name: Martin, Age: 16
Name: Erik, Age: 35
Name: Victor, Age: 46
Name: Grace, Age: 34

you can do it like this too:

foreach syntax: **$something as $key => $value
For each element of the $employeeAges associative array I want to refer to the key as $key and the value as $value.

$employeeAges;
$employeeAges["Lisa"] = "28";
$employeeAges["Martin"] = "16";
$employeeAges["Erik"] = "35";
$employeeAges["Victor"] = "46";
$employeeAges["Grace"] = "34";

foreach( $employeeAges as $name => $age){
    echo "Name: $name, Age: $age <br />";
}

The operator "=>" represents the relationship between a key and value. You can imagine that the key points => to the value. In our example we named the key $key and the value $value. However, it might be easier to think of it as $name and $age. Below our example does this and notice how the output is identical because we only changed the variable names that refer to the keys and values.

You still get the same output.

Output
Name: Lisa, Age: 28
Name: Martin, Age: 16
Name: Erik, Age: 35
Name: Victor, Age: 46
Name: Grace, Age: 34

edited Mar 8, 2016 at 9:26

answered Mar 8, 2016 at 9:20

XsiSecOfficial

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2 Comments

user999 Over a year ago

Thank you, this was very helpful.

XsiSecOfficial Over a year ago

Let me know if you got any questions I may be able to help you =) Regards

Drudge Rajen · Accepted Answer · 2016-03-08 03:46:10Z

1

Something like this :

foreach($results as $result){
$resultData[] = $result['your_value'];
}
print_r($resultData);
echo $resultData[0];

answered Mar 8, 2016 at 3:46

Drudge Rajen

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3 Comments

Drudge Rajen Over a year ago

can you please update your question with the array you get by doing var_dump($results) ?

user999 Over a year ago

Yes you are right, I am trying to get this print_r result. I tried $results = print_r($resultData, true); but this returns nothing, what is the best way to get this result? Thank you very much for your help

naseeba c Over a year ago

@user2983797 try this var_dump($results)

UWU_SANDUN · Accepted Answer · 2016-03-08 04:56:06Z

1

Try with this code

 $resultData= array();`
 foreach($results as $key => $result){
      $resultData[$key] = $result['names'];
 }
  echo "<pre>", print_r($resultData, true);

answered Mar 8, 2016 at 4:56

UWU_SANDUN

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Comments

Maha Dev · Accepted Answer · 2016-03-08 03:46:53Z

0

Here is the proper code. You have not declared the i properly

$resultData= array();
$i = 0;
foreach($results as $result){
      $resultData[$i] = $result['names'];
      $i++
}
//for single
echo $resultData[0];

//for whole
print_r($resultData);

answered Mar 8, 2016 at 3:46

Maha Dev

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6 Comments

GaLaBoOnZ Over a year ago

just asking, why do you need to declare $i, if you use foreach anyway?

Maha Dev Over a year ago

@GaLaBoOnZ , actually i relates the answer with the question. He used variable 'i', that why i told him the correct format of using $i. Nothing else. so many other ways are available but user should understand, what mistake he had done.

GaLaBoOnZ Over a year ago

@MahaDev Oh ok, my bad just noticed that the post was edited and removed $i from the post. :P

Maha Dev Over a year ago

@GaLaBoOnZ, whatever i got -1. :(

GaLaBoOnZ Over a year ago

@MahaDev +1 for explanation :)

|

Amit Chauhan · Accepted Answer · 2016-03-08 04:00:08Z

0

try to use explode in foreach loop

 foreach($results as $result)
 {
      $result = explode(",",$result['names']);
 }
 print_r($result);

if you have a single name in $result then implode it with "," and explode it out side

     foreach($results as $result)
     {
          $val = implode(",",$result['names']);
     }
     $value = explode(","$val);
     print_r($value[0]);

answered Mar 8, 2016 at 4:00

Amit Chauhan

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1 Comment

Amit Chauhan Over a year ago

if you have signle name in $result then use implode of it and explode it in out side

user999 · Accepted Answer · 2016-03-08 16:07:22Z

0

Yes, I got it working, this is the code. Thank you everyone.

      $results;         
      $results[0];
      $results[1];
      $results[2];
       $t = array();
        foreach($results as $key => $value)
        {        
          $t[]= $value['name'];
        }  
        $a = "'".implode("','", $t)."'";
        echo $a;

answered Mar 8, 2016 at 16:07

user999

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