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我的 while 循环是：
我试图插入json_decode并且json_encode每个都没有另一个，但是没有运气。我不断收到错误JSON.parse：JSON 数据后出现意外的非空白字符

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";
}

我知道json_encode必须退出 while 循环，但不明白如何执行此操作。
任何帮助表示赞赏。谢谢你。

4

1 回答 1

1

您引用的特定错误（JSON.parse: unexpected non-whitespace character after JSON data）是 Javascript 对其进行解码时。

json_encode从混合值参数返回一个字符串。json_encode 文档

您要么需要echo输出结果，要么将其添加到另一个变量以稍后输出。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";
}

如果您尝试对多行进行编码并将它们全部返回，则最好将其添加$result到数组中并在while循环外对其进行编码，就像您不这样做一样，您的 JSON 字符串可能如下所示：

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

这不是有效的 JSON，因为它一次只能解析一个对象或数组。下面是一个有效的 JSON 字符串

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

这是您的数据的数组表示，并将解析为您的 JSON 编码对象列表。这是您在回显之前使用数组存储 JSON 的代码。

$results = array();
while ($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);
}

在我上面的示例中，您正在回显有关选择人口或 GDP 的字符串，因为它不是 JSON 编码的，并且上面的部分否则会以 JSON 编码，您在尝试解码时可能会遇到解析错误。如果此 PHP 页面旨在返回 JSON 编码的数据，而您的错误消息不是 JSON 编码的，那么获取该值的任何内容都可能存在问题。

于 2013-04-27T02:28:05.417 回答