Echo array values of column not working

Question

I want to display the values of my array. but instead of displaying:

1509 1510 1511 it display ArrayArrayArray. What does it mean?

include("db_PSIS.php");
$sql_dc = "SELECT Child, Datecode 
                    FROM traveller_merging15
                    WHERE Parent='" . $_REQUEST["txt_traveller_no"] . "'
                    ORDER BY Merge_ID ASC";
$res_dc = mysql_query($sql_dc);

$dcode1 = $row_dc['Datecode'];
$storeArray = array();

if (!$res_dc) {
    echo "No data fetched!";
} else {
    while ($row_dc = mysql_fetch_array($res_dc, MYSQL_ASSOC)) {
        $storeArray[] = $dcode1;
        echo "{$storeArray} <br>";
    }
    //$str_dc=implode(",",$storeArray);
    //echo $str_dc;
}

It means you are trying to echo an array. var_dump($storeArray) and check the value of it!! — Saty
– Saty, Commented Nov 23, 2015 at 6:25
Your this ` $storeArray[] = $dcode1; ` statement says you are storing $dcode1 value in array and you are printing $storeArray variable. As it is array it will display ArrayArrayArray only. Use print_r($storeArray) — Apb
– Apb, Commented Nov 23, 2015 at 6:27
Hi @Saty, i tried to put var_dump($storeArray); And I'm getting this line: array(1) { [0]=> NULL } array(2) { [0]=> NULL [1]=> NULL } array(3) { [0]=> NULL [1]=> NULL [2]=> NULL } — Kimlyn Tormes
– Kimlyn Tormes, Commented Nov 23, 2015 at 6:55
Check my answer below because you are trying to read data before retrieve it from database — Saty
– Saty, Commented Nov 23, 2015 at 6:57

Saty · Accepted Answer · 2015-11-23 06:30:05Z

1

You are assign the value of $row_dc['Datecode']; before fetch data from database. You need to do fetch data inside while loop and echo it

$res_dc = mysql_query($sql_dc);
if (!$res_dc) {
    echo "No data fetched!";
} else {
    while ($row_dc = mysql_fetch_array($res_dc, MYSQL_ASSOC)) {
        echo $row_dc['Datecode'];
    }  
 }

Note:- mysql is deprecated instead use mysqli or PDO

answered Nov 23, 2015 at 6:30

Saty

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Comments

Sugumar Venkatesan · Accepted Answer · 2015-11-23 06:26:46Z

0

   else {
        while ($row_dc = mysql_fetch_array($res_dc, MYSQL_ASSOC)) {
        $storeArray[] =  $dcode1; 
        echo "{$storeArray} <br>";
        }

instead try

   else {
        while ($row_dc = mysql_fetch_array($res_dc, MYSQL_ASSOC)) {
        $storeArray[] =  $dcode1; 
        echo "{$storeArray[0]} <br>";
        }

answered Nov 23, 2015 at 6:26

Sugumar Venkatesan

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Comments

Niranjan N Raju · Accepted Answer · 2015-11-23 06:31:49Z

0

Try this,

else {
  while ($row_dc = mysql_fetch_array($res_dc, MYSQL_ASSOC)) {
  $child =  $row_dc['Child']; 
  $Datecode =  $row_dc['Datecode']; 
  echo "$child <br> $Datecode";
}

if you are getting more that 1 row, use for() loop

else {
  while ($row_dc = mysql_fetch_array($res_dc, MYSQL_ASSOC)) {
  $count = count($row_dc);
  for($i = 0; $i < $count; $i ++){
      $child =  $row_dc[$i]['Child']; 
      $Datecode =  $row_dc[$i]['Datecode']; 
      echo "$child <br> $Datecode";
  }

}

answered Nov 23, 2015 at 6:31

Niranjan N Raju

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Comments

Parth Chavda · Accepted Answer · 2015-11-23 06:33:22Z

0

-first array don't print value using echo

-use print_r($array) to print key value pair(print a array)

your solution is

 $i = 0;
 while ($row_dc = mysql_fetch_array($res_dc, MYSQL_ASSOC)) {
          $storeArray[$i] =  $dcode1; 
          echo $storeArray[$i].'<br>';
          $i++;
  }

NOTE::-- Use single Quote instead of double Quote and mysql is deprecated instead use mysqli or PDO

answered Nov 23, 2015 at 6:33

Parth Chavda

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