我有一个while循环,其中我想将PHP转换为有效的json
我的while循环是:我有一个while循环,其中我想将PHP转换为有效的json,php,encode,json,Php,Encode,Json,我的while循环是: 我试着插入json_decode和json_encode,并且每一个都没有插入另一个,但没有成功。我不断收到错误JSON.parse:JSON数据后出现意外的非空白字符 我知道json\u encode必须在while循环之外,但不知道如何做到这一点。 感谢您的帮助。谢谢。您所引用的特定错误(JSON.parse:JSON数据后意外的非空白字符)是Javascript解码时出现的 json_encode从混合值参数返回字符串 您需要echo输出结果,或者将其添加到另一个变
我试着插入
json_decode
和json_encode
,并且每一个都没有插入另一个,但没有成功。我不断收到错误JSON.parse:JSON数据后出现意外的非空白字符
我知道json\u encode
必须在while循环之外,但不知道如何做到这一点。感谢您的帮助。谢谢。您所引用的特定错误(JSON.parse:JSON数据后意外的非空白字符)是Javascript解码时出现的
json_encode
从混合值参数返回字符串
您需要echo
输出结果,或者将其添加到另一个变量中,以便稍后输出。
例如
如果您试图对多行进行编码并将其全部返回,那么最好将$result
添加到数组中,并在之外对其进行编码,而如果不这样做,则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}]
这是数据的数组表示形式,将解析为JSON编码对象的列表。下面是您的代码,它使用数组在回显JSON之前存储JSON
$results = array();
while ($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,因为这不是JSON编码的,并且上面的部分将以JSON编码,所以在尝试解码时可能会出现解析错误。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,则很可能在获取值时遇到问题。您试图编码什么?您没有提供此类信息。@CurtisCrewe我想对$result
进行编码,如上面的代码所示。当json_encode
在while循环中时,我会得到空白错误,但是当我把它放在外部时,我只得到1个(最后一个)结果,而不是所有结果。
[{gold:123,silver:456,bronze:789},{gold:987,silver:654,bronze:321}]
$results = array();
while ($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);
}