dongyi3776 2018-08-13 07:09
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PHP / MySQL:SyntaxError:JSON.parse:JSON数据第1行第1列的意外字符

I have this MySQL table:

mysql> select * from  members;
+-------+-----------+-----------+
| memid | firstname | lastname  |
+-------+-----------+-----------+
|     1 | billal    | begueradj |
|     2 | bill      | gates     |
|     3 | steve     | jobs      |
+-------+-----------+-----------+
3 rows in set (0.00 sec)

I have this code:

<?php
$output = array('error' => false);
$members = array();

try {
    $db = new PDO('mysql:host=localhost;dbname=bill;charset=utf8',
                   'root',
                   ''
    );
} catch(Exception $e) {
    die('Error in connecting to DB: <br/>'.$e->getMessage());
}

$response = $db->query('SELECT * FROM members');
while($row = $response->fetch()){
    echo $row['firstname'].' ';
    echo $row['lastname'].'<br/>';
    array_push($members, $row);
}

$output['members'] = $members;
$response->closeCursor();
$json = json_encode($out);
echo $json; // outputs correctly
header("Content-type: application/json");    // error here
die();
?>

And I am getting this error message when I run the PHP file containing the PHP code above:

SyntaxError: JSON.parse: unexpected character at line 1 column 1 of the JSON data

Why this happens?

P.S. Of course, when I comment this line: //header("Content-type: application/json"); the error message disappears

EDIT: New code version following the comments below:

<?php
    header("Content-type: application/json");
    $output = array('error' => false);
    $members = array();

    try {
        $db = new PDO('mysql:host=localhost;dbname=bill;charset=utf8',
                       'root',
                       ''
        );
    } catch(Exception $e) {
        die('Error in connecting to DB: <br/>'.$e->getMessage());
    }

    $response = $db->query('SELECT * FROM members');
    while($row = $response->fetch()){       
        array_push($members, $row);
    }

    $output['members'] = $members;
    $response->closeCursor();
    $json = json_encode($out); 
    //echo $json;    

?>

Still getting the same error message

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2条回答 默认 最新

  • douti6740 2018-08-13 07:20
    关注

    header() should be placed above all output, so before you echo anything.

    Plus, since your not encoding the whole thing to JSON, instead only parts, I'm guessing your missing start and end of the JSON.

    Correct JSON outputs:

    ex1:

    {
      "key1": "value1",
      "key2": "value2",
      "key3": "value3"
    }
    

    ex2:

    [
      {
        "key1": "value1",
        "key2": "value2",
        "key3": "value3"
      },
      {
        "key1": "value1",
        "key2": "value2",
        "key3": "value3"
      },
      {
        "key1": "value1",
        "key2": "value2",
        "key3": "value3"
      }
    ]
    

    PHP:

    <?php
    
    $output = ["error" => false, "members" => []];
    
    try {
      $db = new PDO("mysql:host=localhost;dbname=bill;charset=utf8", "root", "");
    } catch(Exception $e) {
      die("Error in connecting to DB: <br/>{$e->getMessage()}");
    }
    
    $response = $db->query("SELECT * FROM members");
    
    while($row = $response->fetch(PDO::FETCH_ASSOC)) {
      array_push($output["members"], $row);
    }
    
    $response->closeCursor();
    
    $json = json_encode($output);
    
    header("Content-type: application/json");
    
    echo $json;
    
    本回答被题主选为最佳回答 , 对您是否有帮助呢?
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