duanke3985 2014-07-23 02:35
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mysqli_fetch_array()期望参数1为mysqli_result,在某些列中选中

I'm trying to select all of my column and it works perfectly with this

$query = "SELECT * FROM employee";
$result = mysqli_query($con, $query);

while ($employee = mysqli_fetch_array($result)) {
    # code...
    print (json_encode($employee));
}

and I also succeed select some of my column using this code

$query = "SELECT ID, NAME, LAT FROM employee";
$result = mysqli_query($con, $query);

but I failed when I try to select LONG column, it said that

mysqli_fetch_array() expects parameter 1 to be mysqli_result

It just show when I select LONG column, but it works well if I select another column.

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  • dq13733519262 2014-07-23 02:37
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    "But i failed when i try to select LONG column"

    That's because LONG is a MySQL reserved word.

    Either wrap it in backticks, or choose another word for it.

    I.e.:

    $query = "SELECT ID, NAME, `LONG` FROM employee";
    

    or

    $query = "SELECT ID, NAME, LAT, `LONG` FROM employee";
    
    本回答被题主选为最佳回答 , 对您是否有帮助呢?
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