我有一個while循環和裏面我想PHP轉換爲有效的JSON

Question

我while循環：
我試圖插入json_decode和json_encode也是他們中的每一個沒有其他的，但有沒有運氣。我不斷收到錯誤JSON.parse：後JSON數據意外非空白字符

while ($row = $res->fetchRow()){ 
    $resGold = $row['gold']; 
    $resSilver = $row['silver']; 
    $resBronze = $row['bronze']; 
    $resGdp = $row['gdp']; 
    $resPopulation = $row['population']; 
    $resCountry = $row['country_name']; 
    $gold_score = ($resGold * $gold_value); 
    $silver_score = ($resSilver * $silver_value); 
    $bronze_score = ($resBronze * $bronze_value); 
    $score_pop = (($gold_score + $silver_score + $bronze_score)/$resPopulation); 
    $score_gdp = (($gold_score + $silver_score + $bronze_score)/$resGdp); 
    if($population == 'true'){ 
     $result = $res->fetchRow(); 
     $result['score'] = "$score_pop"; 
     json_encode($result); 
    } 
    else if($gdp == 'true'){ 
     $result = $res->fetchRow(); 
     $result['score'] = "$score_gdp"; 
     json_encode($result); 
    } 
} 
if($population == 'false' && $gdp == 'false'){ 
    echo "Please select either population or gdp from view.htm"; 
} 

我明白json_encode必須跳出while循環，但不知道如何做到這一點。
任何幫助表示讚賞。謝謝。

Answer 1

您引用的特定錯誤（JSON.parse：JSON數據後的意外非空白字符）是Javascript正在解碼它時。

json_encode從混合值參數中返回一個字符串。 json_encode Doc

您可能需要echo以外的結果或將其添加到另一個變量稍後輸出。 例如。

while ($row = $res->fetchRow()){ 
    $resGold = $row['gold']; 
    $resSilver = $row['silver']; 
    $resBronze = $row['bronze']; 
    $resGdp = $row['gdp']; 
    $resPopulation = $row['population']; 
    $resCountry = $row['country_name']; 
    $gold_score = ($resGold * $gold_value); 
    $silver_score = ($resSilver * $silver_value); 
    $bronze_score = ($resBronze * $bronze_value); 
    $score_pop = (($gold_score + $silver_score + $bronze_score)/$resPopulation); 
    $score_gdp = (($gold_score + $silver_score + $bronze_score)/$resGdp); 
    if($population == 'true'){ 
     $result = $res->fetchRow(); 
     $result['score'] = "$score_pop"; 
     echo json_encode($result); 
    } 
    else if($gdp == 'true'){ 
     $result = $res->fetchRow(); 
     $result['score'] = "$score_gdp"; 
     echo json_encode($result); 
    } 
} 
if($population == 'false' && $gdp == 'false'){ 
    echo "Please select either population or gdp from view.htm"; 
} 

如果你想編碼多行，並返回所有這些，你會過得更好添加$result到一個數組和編碼的while循環外，如果你不這樣做，你的JSON字符串可以看看如：

{gold:123,silver:456,bronze:789}{gold:987,silver:654,bronze:321} 

這是無效的JSON，因爲它只能一次解析一個對象或數組。下面是一個有效的JSON字符串

[{gold:123,silver:456,bronze:789},{gold:987,silver:654,bronze:321}] 

這是數據的數組表示形式，將解析爲您的JSON編碼對象的列表。這是您的代碼在使用數組來存儲JSON之前對其進行回顯。

$results = array(); 
while ($row = $res->fetchRow()){ 
    $resGold = $row['gold']; 
    $resSilver = $row['silver']; 
    $resBronze = $row['bronze']; 
    $resGdp = $row['gdp']; 
    $resPopulation = $row['population']; 
    $resCountry = $row['country_name']; 
    $gold_score = ($resGold * $gold_value); 
    $silver_score = ($resSilver * $silver_value); 
    $bronze_score = ($resBronze * $bronze_value); 
    $score_pop = (($gold_score + $silver_score + $bronze_score)/$resPopulation); 
    $score_gdp = (($gold_score + $silver_score + $bronze_score)/$resGdp); 
    if($population == 'true'){ 
     $result = $res->fetchRow(); 
     $result['score'] = "$score_pop"; 
     array_push($results,$result); 
    } 
    else if($gdp == 'true'){ 
     $result = $res->fetchRow(); 
     $result['score'] = "$score_gdp"; 
     array_push($results,$result); 
    } 
} 
if($population == 'false' && $gdp == 'false'){ 
    echo "Please select either population or gdp from view.htm"; 
} 
else 
{ 
    //Note: This is in the 'else' statement as echoing the JSON then that string 
    //  will also cause errors as it ends up not being valid JSON anymore 
    echo json_encode($results); 
} 

正如我在上面的例子，你是呼應字符串就選擇人口或GDP，因爲這是不JSON編碼及以上，否則在JSON編碼的部分，你可以嘗試解碼時遇到解析錯誤它。如果此PHP頁面旨在返回JSON編碼數據，並且您的錯誤消息不是JSON編碼，那麼無論獲取值是什麼，都可能會出現問題。