我正在尝试制作一个php web服务,它可以从我在学校的网络空间上的MySQL数据库中获取数据,并将数据解析为JSON。然后我从jQuery调用这个php函数,但php似乎返回了一个空数组。
以下是php代码:
<?php
// Create connection
$con=mysqli_connect("mysqlstudent","sylvainvansteela","zei8quea0eeP","sylvainvansteela");
if (mysqli_connect_errno($con))
{
echo "Failed to connect to MySQL: " . mysqli_connect_error();
} else {
$mysqlstring = "SELECT * FROM customers";
$result = mysqli_query($con,$mysqlstring);
$rows = array();
while($r = mysql_fetch_array($result)) {
$rows["id"] = $r[0];
$rows["email"] = $r[1];
}
header("Content-type: application/json");
echo json_encode($rows);
}
?>
这是jQuery代码:
函数getCustomers(){
var url = 'http://student.howest.be/sylvain.vansteelandt/fedex/server/getcustomers.php';
$.getJSON(url,function(data){
if(data){
alert(data);
console.log(data.length);
} else {
alert('error');
}
});
};
getCustomers();
我建议更改PHP的以下部分:
$rows = array();
$i = 0; // add this
while($r = mysqli_fetch_array($result)) { // mysqli!
$rows[$i]["id"] = $r[0];
$rows[$i]["email"] = $r[1];
$i++; // don't forget to increment
}
您需要在此处进行更改:
$inc = 0;
while($r = mysqli_fetch_array($result)) { //here you are using the mysql it should be mysqli
$rows[$inc]["id"] = $r[0];
$rows[$inc]["email"] = $r[1];
$inc++;
}
如果可能,则使用mysqli_fetch_assoc
。
$rows = array();
while($r = mysqli_fetch_array($result)) {
$row = array();
$row['id'] = $r[0];
$row['email'] = $r[1];
array_push($rows, $row);
}
echo json_encode($rows);
// OR simplify
$rows = array();
while($r = mysqli_fetch_array($result)) {
array_push($rows, array('id'=>$r[0], 'email'=>$r[1]));
}
echo json_encode($rows);