我已经在php中创建了服务，但我遇到了一些问题，这里是我实现了我的代码
<?php
include("db.php");
$month=10;
$year=date("Y");
$sq=mysql_query('select * from city where id=1 ORDER BY city ASC');
while($re1=mysql_fetch_array($sq)){
$id = $re1['id'];
$city = $re1['city'];
$city_id = $re1['id'];
echo '{ <br>"'.$city.'" : [ <br>';
$sql='select * from price where month="'.$month.'" and year="'.$year.'" and city="'.$id.'" ORDER BY day ASC';
$sql2=mysql_query($sql);
while($re2=mysql_fetch_array($sql2)){
echo '{ <br>';
echo '"date" : "'.$re2["day"].'-'.$re2["month"].'-'.$re2["year"].'",<br>';
if($re2['price']==''){ echo "Price : -- , <br>";}
else{
echo '"Price" : "'.$re2["price"].'" <br>';
}
echo '}, <br>';
}
echo ']<br>}';
}
?>
这样的响应json格式。
{
"Ahmedabad" : [
{
"date" : "1-10-2014",
"Price" : "353"
},
{
"date" : "2-10-2014",
"Price" : "353"
},
{
"date" : "3-10-2014",
"Price" : "327"
},
]
}
但是在json格式结束字典中如何删除此,。你能建议我吗。
答案 0 :(得分：1)
首先，您不能在JSON响应中拥有<br>，除非此响应仅用于显示而非实际使用。
为了在最后删除,，您需要每个＆＃34;行＆＃34;在您可以编辑的变量中创建，而不仅仅是回显该行。
以下是一些可以实现这一目标的代码：
<?php
include("db.php");
$month=10;
$year=date("Y");
$sq=mysql_query('select * from city where id=1 ORDER BY city ASC');
while($re1=mysql_fetch_array($sq)){
$id = $re1['id'];
$city = $re1['city'];
$city_id = $re1['id'];
echo '{
"'.$city.'" : [
';
$sql='select * from price where month="'.$month.'" and year="'.$year.'" and city="'.$id.'" ORDER BY day ASC';
$sql2=mysql_query($sql);
while($re2=mysql_fetch_array($sql2)){
$line = '{
';
$line.= '"date" : "'.$re2["day"].'-'.$re2["month"].'-'.$re2["year"].'",
';
if($re2['price']=='')
{
$line.= "Price : -- ,
";
}
else
{
$line.= '"Price" : "'.$re2["price"].'"
';
}
$line.= '},';
echo $line;
}
$line = substr($line,0,-1);
echo ']
}';
}
?>
您应该使用PHP内置的JSON格式化函数，如json_encode()
为了做出良好的JSON响应，您需要正确处理数据以避免格式化错误。
所以这段代码更好地利用了PHP和MySQL：
<?php
include("db.php");
$month=10;
$year=date("Y");
$sq=mysql_query('select * from city where id=1 ORDER BY city ASC');
while($re1=mysql_fetch_array($sq)){
$id = $re1['id'];
$city = $re1['city'];
$city_id = $re1['id'];
$sql='select * from price where month="'.$month.'" and year="'.$year.'" and city="'.$id.'" ORDER BY day ASC';
$sql2=mysql_query($sql);
while($re2=mysql_fetch_array($sql2)){
// create each line in the city
$line["date"] = $re2["day"].'-'.$re2["month"].'-'.$re2["year"];
if($re2['price']=='')
{
$line["Price"] = "--";
}
else
{
$line["Price"] = $re2["price"];
}
// add the line to the city
$cities[$city][] = $line;
}
}
// encode to json
json_encode($cities);
?>
答案 1 :(得分：0)
我认为应该这样做。我现在在日期/价格块的开头添加一个逗号，每个日期/价格组合除了第一个（由新的$ ct变量检查）。
<?php
include("db.php");
$month=10;
$year=date("Y");
$sq=mysql_query('select * from city where id=1 ORDER BY city ASC');
while($re1=mysql_fetch_array($sq)){
$id = $re1['id'];
$city = $re1['city'];
$city_id = $re1['id'];
echo '{ <br>"'.$city.'" : [ <br>';
$sql='select * from price where month="'.$month.'" and year="'.$year.'" and city="'.$id.'" ORDER BY day ASC';
$sql2=mysql_query($sql);
$ct = 0;
while($re2=mysql_fetch_array($sql2)){
if ($ct > 0) { echo ', ';}
ct++;
echo '{ <br>';
echo '"date" : "'.$re2["day"].'-'.$re2["month"].'-'.$re2["year"].'",<br>';
if($re2['price']==''){ echo "Price : -- , <br>";}
else{
echo '"Price" : "'.$re2["price"].'" <br>';
}
echo '}<br>';
}
echo ']<br>}';
}
?>
答案 2 :(得分：0)
试试这个
include("db.php");
$month=10;
$year=date("Y");
$sq=mysql_query('select * from city where id=1 ORDER BY city ASC');
$city=array();
while($re1=mysql_fetch_array($sq)){
$city['city'] = $re1['city'];
$sql='select * from price where month="'.$month.'" and year="'.$year.'" and city="'.$id.'" ORDER BY day ASC';
$sql2=mysql_query($sql);
while($re2=mysql_fetch_array($sql2)){
$city['city'] = array("date" => $re2["day"].'-'.$re2["month"].'-'.$re2["year"];
if($re2['price']==''){
$city['city'] = array('price'=>'--');
}else{
$city['city'] = array('price'=>$re2["price"]);
}
}
}
echo json_encode($city);
?>
答案 3 :(得分：0)
欢迎使用Stack Overflow。而不是回答你的问题，确切地说要修改代码的内容，这里有一个重构版本的代码，其中包含我在PHP开始编程时会喜欢的注释。希望有适合你的东西：
$month = 10;
$year = date('Y');
// 1. You can get the data you're after in a single query using a LEFT JOIN.
// In your original code you run one database query for every city, which
// quickly slows down the program as you add more cities. This change is
// mostly for keeping the code clean in this case, but in practice it's
// good to avoid unnecessary back and forth with the database.
//
// 2. The MySQL DATE data type opens up a lot of functionality in your queries
// that isn't available when storing day, month and year in separate columns.
//
// 3. The mysql_* functions are deprecated and no longer exist in PHP 5.5.0,
// so you ideally shouldn't be using them (see the red box in the docs):
//
// http://php.net/manual/en/function.mysql-query.php
//
// I've opted to demo PDO with a parameterised query as I like it, but you
// could use the mysqli_* functions instead. The following pretends that
// $connection is an instance of PDO. The variable `$stmt` stands for
// 'statement'. Further info on PDO:
//
// http://php.net/manual/en/class.pdo.php
// http://wiki.hashphp.org/PDO_Tutorial_for_MySQL_Developers
//
$stmt = $connection->prepare(
'SELECT
city.name AS city,
price.price,
DATE_FORMAT(price.date, '%d-%m-%Y') AS date
FROM city
LEFT JOIN price ON price.city_id = city.id
WHERE MONTH(price.date) = :month AND YEAR(price.date) = :year
ORDER BY DAY(price.date) ASC'
);
$stmt->bindValue(':month', $month);
$stmt->bindValue(':year', $year);
$stmt->execute();
$results = $stmt->fetchAll(PDO::FETCH_ASSOC);
// This short array syntax is valid in PHP 5.4 and higher.
// If you're using 5.3 or lower, `[]` would be `array()`
$output = [];
foreach ($results as $row) {
$output[$row['city']] = [
'date' => $row['date'],
'price' => $row['price'] ?: '--'
];
}
json_encode($output);
一般说明：
json_encode而不是手动构建JSON字符串）。如果您尝试做的事情看起来过于困难，可能有更好的方法来解决问题。