如何在while循环前检查php mysqli_fetch_array是否为空

Question

我想在运行while循环之前确保有结果，但我尝试的所有方法似乎都删除了第一个结果。

$nearbyResult = mysqli_query($con,$sqlNearby);

if(mysqli_fetch_array($nearbyResult) == 0) {
    echo '<p>No results found, <a href="/add.php">Add your property here</a>.</p>';
} else {
    while($rowNearby = mysqli_fetch_array($nearbyResult)) {
    }
}

Answer 1

这一行将获取结果集的第一行并将其丢弃到bin中：

if(mysqli_fetch_array($nearbyResult) == 0) {

更改为：

if( ! mysqli_num_rows($nearbyResult) ) {

并检查你的异常函数的返回：

if( ! $nearbyResult = mysqli_query($con,$sqlNearby) ) {
    echo "Mysql error: " . mysqli_error($con);
}

Answer 2

可以使用mysql_row_count方法计算查询http://php.net/manual/en/mysqli-result.num-rows.php中返回的行数

Answer 3

尝试将结果赋给一个变量，作为if语句的一部分。如果mysqli_fetch_array()没有可处理的结果集，它将返回false。

if($rowNearby = mysqli_fetch_array($nearbyResult)) {
    //There was a result, work with it here
    doStuffWith($rowNearby);
} else {
    //No records in your result set, handle as desired
}