我写了这个代码:
require("config.php");
$rows = array();
$query = "SELECT accountname, bill_city, bill_code, bill_country, bill_street, latitude, longitude, setype FROM vtiger_accountbillads, vtiger_account, vtiger_geocoding WHERE accountaddressid = accountid AND accountid = crmid";
$result = mysqli_query($connection, $query);
$rows_number = $result->num_rows;
echo $rows_number . "<br/><br/>";
$row = mysqli_fetch_array($result);
if($result->num_rows > 0){
for($i=0; $i < $rows_number; $i++) {
$rows[] = array("name" => $row[0],
"city" => $row[1],
"code" => $row[2],
"country" => $row[3],
"street" => $row[4],
"latitude" => $row[5],
"longitude" => $row[6],
"type" => $row[7]);
}
}
$json = json_encode($rows);
print $json;
mysqli_free_result($row);
mysqli_close($connection);
我正试图使用上面代码中编写的查询获取几个数据,但它显示了47次第一行。为什么?我做错了什么?谢谢
您需要遍历MySQL返回的结果集。这意味着为该结果集的每一行调用mysqli_fetch_array()
。您可以使用while
循环:
while($row = mysqli_fetch_assoc($result)) {
$rows[] = array("name" => $row['name'],
"city" => $row['bill_city'],
"code" => $row['bill_code'],
"country" => $row['bill_country'],
"street" => $row['bill_street'],
"latitude" => $row['latitude'],
"longitude" => $row['longitude'],
"type" => $row['setype']);
}
}
不要使用for loop
来获取SQL查询。改为使用while
:
if($result->num_rows > 0){
while($row = mysqli_fetch_assoc($result))
{
$rows[] = array("name" => $row[0],
"city" => $row[1],
"code" => $row[2],
"country" => $row[3],
"street" => $row[4],
"latitude" => $row[5],
"longitude" => $row[6],
"type" => $row[7]);
}
}