2017-06-19 44 views
-1

我有一個PHP文件,給了我需要作爲一個字符串的響應,我需要它是一個數組編碼在JSON中,即時通訊不是很好與PHP,可以任何一個幫幫我 ? 這是它如何工作的罰款作爲字符串 - printf("user: \"%s\" \"%s\" email: \"%s\" \n", $row['firstname'], $row['lastname'],$row['email']);php文件,返回一個字符串JSON編碼沒有工作

但我需要它作爲JSON數組這就是我試圖做到這一點 - array_push($mynewArray,array("firstname \%s"=>$row['firstname'],"lastname \%s"=>$row['lastname'],"email \%s"=>$row['email'])); echo json_encode($mynewArray);

this is the screen shot that shows how i tried to do it

+0

使用表單'的數組鍵姓名\%s'可能不會做出最有意義... – CBroe

回答

0

我想你可以這樣嘗試,假設記錄集生成正確,並在var 0中可用

$mynewArray=array(); 
foreach($result as $row){ 
    $mynewArray[]=$row; 
} 
echo json_encode($mynewArray); 


Or 

$mynewArray=array(); 
foreach($result as $row){ 
    $mynewArray[]=array(
     'firstname'=>$row['firstname'], 
     'lastname'=>$row['lastname'], 
     'email'=>$row['email'], 
    ); 
} 
echo json_encode($mynewArray); 
+0

你救了我thannks你RamRaider – raddaoui

0

這可能是什麼你在之後:

<?php 
$cluster = Cassandra::cluster() 
       ->build(); 
$keyspace = 'msata'; 
$session = $cluster->connect($keyspace); 

$result = $session->execute(new Cassandra\SimpleStatement 
      ("SELECT * FROM msata.users") 
     ); 

// Create the variable array here so it's used correctly in the loop 
$mynewArray = array(); 

foreach($result as $row){ 

    // Create an array for each individual user 
    $user_array = array('firstname' => $row['firstname'], 'lastname' => $row['lastname'], 'email' => $row['lastname']); 

    // Add the array to the main array 
    $mynewArray = array_push($mynewArray, $user_array); 

} 

echo json_encode($mynewArray);