求助！！php json 字符串问题-php教程-PHP中文網

求助！！php json 字符串问题

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發布： 2016-06-23 14:39:15

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本帖最后由 zl2727 于 2013-12-10 16:14:43 编辑

$proquery ="select * from #@__keyproduct where typeid='".$id."' order by corank asc";

$dsql->SetQuery($proquery);
$dsql->Execute();
$i=0;
while($row = $dsql->GetArray()) {

$arrystr.='{"id":"'.$row["id"].'","title":"'.$row["goodstitle"].'","goodsTitle":"'.$row["goodstitle"].'","price":"'.$row

["price"].'","imgUrl":"'.$row["imgurl"].'","goodsUrl":"'.$row["goodsurl"].'","brandName":"'.$row

["brandname"].'","brandUrl":"'.$row["brandurl"].'"},';

$i++;
}

if($i>0){ $arrystr=substr($arrystr,0,$arrystr.lenght-1); }
如果这样写
$a='{"status":"1","result":['.$arrystr.'],"moreUrl":"'.$moreurl.'"}';
AJAX返回 {"status":"1","result":[],"moreUrl":"http:\/\/"}
result 是空的

如果下面这样写就是正常
//$a='{"status":"1","result":[{"id":"1","title":"补水润肤膏真的很好很补水皮肤有弹性","goodsTitle":"补水润肤膏真的很好很补水皮肤有弹","price":"59.9","imgUrl":"http:///sadas213/","goodsUrl":"http:///uploads/allimg/c131205/

13V23912150-114L_lit.jpg","brandName":"","brandUrl":"http:///dsadasdas/"},{"id":"2","title":"补水润肤

222","goodsTitle":"补水润肤

222","price":"59.9","imgUrl":"http:///sadas213/","goodsUrl":"http:///uploads/allimg/c131205

/13V23912150-

114L_lit.jpg","brandName":"","brandUrl":"http:///dsadasdas/"}],"moreUrl":"http://jianfei.ecoo.com.cn/shuig

uo/"}';

$a = iconv("gbk", "UTF-8", $a);
$a=json_decode($a, true);
echo json_encode($a);

请大神们指教一下谢谢！

回复讨论(解决方案)

$r['a'] = 123;echo json_encode($r); //得到 {"a":123}$a[] = $r;$a[] = $r;$a[] = $r;echo json_encode($a); //得到 [{"a":123},{"a":123},{"a":123}]

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你应该知道怎么写了吧？

虽然 php 的 json 函数只支持 utf-8 编码，但你可以要求数据库返回 utf-8 编码的数据

版主您好我将代码修改成这样可还是不行我是学.net的 Php实在太差麻烦您在指导一下谢谢
$a=array();
while($row = $dsql->GetArray()) {
$a[]= $row;
}
$a='{"status":"1","result":'.json_encode($a).',"moreUrl":"'.$moreurl.'"}';
$a = iconv("gbk", "UTF-8", $a);
$a=json_decode($a, true);
echo json_encode($a);

AJAX返回 {"status":"1","result":[],"moreUrl":"http:\/\/"}
result 是空的

1、你用的是什么数据库？不会是 Access 吧？
2、如果你无法让数据库返回 utf-8 编码的数据，可以这样

function gtou($s) { return iconv('gbk', 'utf-8', $s); }$a=array();while($row = $dsql->GetArray()) {	  $a[]= array_map('gtou', $row); }

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3、这样构造输出数据

$res = array(  'status' => '1',  'result' => $a,  'moreUrl' => $moreurl,);echo join_encode($res);

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版主您好这样写输出也是对的
但
AJAX返回 {"status":"1","result":[],"moreUrl":"http:\/\/"}
result 是空的

我的QQ 272718467 要不您加下我的QQ指导一下或者关注我一下可以吗？

这样直接返回数据库读出的数据 AJAX 就可以接到 result
//$a='{"status":"1","result":[{"id":"1","title":"补水润肤膏真的很好很补水皮肤有弹性","goodsTitle":"补水润肤膏真的很好
很补水皮肤有弹
性","price":"59.9","imgUrl":"http:///sadas213/","goodsUrl":"http:///uploads/allimg/c131205/
13V23912150-114L_lit.jpg","brandName":"","brandUrl":"http:///dsadasdas/"},{"id":"2","title":"补水润肤
222","goodsTitle":"补水润肤
222","price":"59.9","imgUrl":"http:///sadas213/","goodsUrl":"http:///uploads/allimg/c131205
/13V23912150-114L_lit.jpg","brandName":"","brandUrl":"http:///dsadasdas/"}],"moreUrl":"http://shuig
uo/"}';

我是mysql数据库

	$JsonStr='[		{"id":"1","name":"\u5f20\u96ea\u6885","age":"27","subject":"\u8ba1\u7b97\u673a\u79d1\u5b66\u4e0e\u6280\u672f"},		{"id":"2","name":"\u5f20\u6c9b\u9716","age":"21","subject":"\u8f6f\u4ef6\u5de5\u7a0b"}	]';	//被解析的字符串只能用?引?,而不能用?引?	$Json=json_decode($JsonStr,TRUE);	echo '<pre class="brush:php;toolbar:false">',var_dump($Json),'

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'; $array=array('a'=>'A','B'=>2); $a=json_encode($array); echo '