mysqli fetch array displays only the first row(mysqli fetch 数组只显示第一行)
本文介绍了mysqli fetch 数组只显示第一行的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着小编来一起学习吧!
问题描述
我写了这段代码:
I have written this code:
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($connection);
我正在尝试使用上面代码中编写的查询来获取多个数据,但它显示了第一行 47 次.为什么?我究竟做错了什么?谢谢!
I'm trying to fetch several data using the query written in the code above, but it displays the first row 47 times. Why? What am I doing wrong? Thanks!
推荐答案
需要遍历 MySQL 返回的结果集.这意味着为该结果集的每一行调用 mysqli_fetch_array()
.你可以使用 while
循环来做到这一点:
You need to iterate through the result set returned by MySQL. That means calling mysqli_fetch_array()
for each row of that result set. You can do that using a while
loop:
while($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']);
}
}
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本文标题为:mysqli fetch 数组只显示第一行


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