ON senses.synsetid = synsets.synsetid
where senses.wordid = 79459我试过CREATE TABLE `synsets` (
`synsetid` int(10) unsigned NOT NULL DEFAULT '0',
`pos` enum('n','v','a','r','s') NOT NULL,
`definition` mediumtext,
PRIMARY KEY (`synsetid`)
) ENGINE=MyISAM DEFAULT CHARSET=utf8;
CREATE TABLE `words` (
`wordid` int(10) unsigned NOT NULL DEFAULT '0',
`lemma` varchar(80) NOT NULL,
`mantiq` varchar(255) NOT NULL DEFAULT '',
PRIMARY KEY (`wordid`),
UNIQUE KEY `unq_words_lemma` (`lemma`)
) ENGINE=MyISAM DEFAULT CHARSET=utf8;
CREATE TABLE `senses` (
`wordid` int(10) unsigned NOT NULL DEFAULT '0',
`synsetid` int(10) unsigned NOT NULL DEFAULT '0',
`senseid` int(10) unsigned DEFAULT NULL,
`sensekey` varchar(100) DEFAULT NULL,
PRIMARY KEY (`wordid`,`synsetid`)
) ENGINE=MyISAM DEFAULT CHARSET=utf8;
CREATE TABLE `samples` (
`synsetid` int(10) unsigned NOT NULL DEFAULT '0',
`sampleid` smallint(5) unsigned NOT NULL DEFAULT '0',
`sample` mediumtext NOT NULL,
PRIMARY KEY (`synsetid`,`sampleid`)
) ENGINE=MyISAM DEFAULT CHARSET=utf8;
但是没有用;
ERROR
$files = array();
$dir = opendir(dirname(realpath(__FILE__)).'/files/'.$fil_fi_row['file_name'].'/');
while ($file = readdir($dir)) {
if ($file == '.' || $file == '..') {
continue;
}
$allfiles[] = $file;
}
$total_arr=array_merge((array)$id,(array)$files);
sort($total_arr);
foreach($total_arr as $key=>$file){
if($key != count($total_arr)-1)
$array[‘Files’][] = array(
‘files’ => $file,
‘id’ => $id,
);
}
}
echo json_encode($array, JSON_UNESCAPED_UNICODE);
当我使用
时JSON_UNESCAPED_UNICODE阿拉伯文件名显示为：
Notice: Use of undefined constant JSON_UNESCAPED_UNICODE - assumed 'JSON_UNESCAPED_UNICODE' in /home/...
Warning: json_encode() expects parameter 2 to be long, string given in /home/..
答案 0 :(得分：0)
试试这个（PHP5.4.0 +）
json_encode($result, JSON_UNESCAPED_UNICODE);
答案 1 :(得分：0)
您需要添加以下代码
header('Content-Type: applcation/json; charset = utf-8');
mysqli_set_charset($conn,"utf8");
为我工作