Question

我正在尝试在执行MySQL查询时将字符串输出到html。它似乎工作，当它回应它它给我错误。代码是：

$result = mysqli_query($con,'SELECT * FROM vehicle_type ORDER BY cat_id desc');

while($row = mysqli_fetch_array($result)) {
$category = $row['category'];
$count++;



$vehicle_types .= "<a href='#' onclick='document.getElementById('vehicle_types').value='$count';$('#exec').show();$('#coach').hide();$('#minibus').hide();$('#limos').hide();vehicle_type_name.innerHTML = 'Vehicle type selected : Executive vehicle';vehicle_selected.innerHTML = ''; ' class='btn new btn-primary'>$category</a>";

}

当我回应结果时，它给了我：

 <a href='#' onclick='document.getElementById('vehicle_types').value='1';$('#exec').show();$('#coach').hide();$('#minibus').hide();$('#limos').hide();vehicle_type_name.innerHTML = 'Vehicle type selected : Executive vehicle';vehicle_selected.innerHTML = ''; ' class='btn new btn-primary'>Minibus</a>

什么时候应该给我：

<a href="#" onclick="document.getElementById('vehicle_type').value='1';$('#exec').show();$('#coach').hide();$('#minibus').hide();$('#limos').hide();vehicle_type_name.innerHTML = 'Vehicle type selected : Executive vehicle';vehicle_selected.innerHTML = ''; " class="btn new btn-primary">Executive cars</a>

由于

Answer 1

我建议将所有onClick语句移到函数中以提高可读性。

此外，与您使用jQuery一致。

$vehicle_types .= '<a href="#" onclick="vehicleType_onClick(\'' . $count . '\')" 
                   class="btn new btn-primary">' . $category . '</a>';

<script>
function vehicleType_onClick(count){
    $('#vehicle_types').val(count);

    $('#exec').show();
    $('#coach').hide();
    $('#minibus').hide();
    $('#limos').hide();
    $('input[name=vehicle_type_name']).html('Vehicle type selected : Executive vehicle');
    $('input[name=vehicle_selected']).html(''); 
}
</script>

将mysql结果返回到php字符串

1 个答案: