我想使用phonegap使应用程序。我试图做一个选择使用PHP返回结果到ajax函数并打印结果...但我不知道我是否正确,我需要把我的HTML代码放在显示结果。试图从php接收json和ajax并打印结果在HTML
Ajax代码:
$(document).ready(function() {
$.ajax({
url : 'http://ip/again/www/index.php',
dataType : "json",
success : function(data){
var html = "";
for($i=0; $i < data.length; $i++){
html += "<strong>Nome:</strong> "+data[$i].nome +" "+
data[$i].sobreNome;
html += " <strong>Cidade:</strong> "+data[$i].cidade;
html += "<br />";
}
$('body').html(html);
}
});
});
</script>
PHP代码:
<?php
$hostname = 'localhost';
$username = 'root';
$password = 'pass';
$database = 'mydb';
try {
$pdo = new PDO("mysql:host=$hostname;dbname=$database", $username,
$password, array(PDO::MYSQL_ATTR_INIT_COMMAND => "SET NAMES utf8"));
//echo 'Conexao efetuada com sucesso!';
}
catch(PDOException $e)
{
echo $e->getMessage();
}
$sql = 'SELECT * FROM incidente ORDER BY codigo' ;
try {
$recebeConexao = $pdo->prepare($sql);
$recebeConexao->execute();
$result = $recebeConexao->fetchAll();
if (count($result)) {
foreach($result as $row) {
$incid=$row;
echo (json_encode($incid));
}
} else {
$incid=0;
}
} catch (PDOException $ex) {
echo "Erro no Banco";
}
?>
有什么问题或错误你是目前正在看?您的帖子中没有足够的信息来帮助您。你的AJAX请求返回了吗?如果是的话,答复的内容是什么? – thatidiotguy
我的AJAX没有返回任何响应...我的PHP返回类似于:{“codigo”:“61”,“0”:“61”,“nome”:“kelly”,“1”:“kelly”, “sobrenome”:“kinder”,“2”:“kinder”,“cidade”:“rio”,“3”:“rio”} {“codigo”:“62”,“0”:“62” “nome”:“jack”,“1”:“jack”,“sobrenome”:“jones”,“2”:“jones”,“cidade”:“bh”,“3”:“bh”} –