我在这里遇到了一个非常非常奇怪的问题。我在 php 中的数组中有数据。我希望数组必须查询数据库并返回结果的每个项目。它只向我显示具体的结果,而不是所有结果。
我的代码:
foreach($my as $k=> $v){
//echo "Key: ". $k . " Value: " . $v . "<br/>";
$sql2 = "SELECT column10 FROM `table` WHERE column1 = '$v' ";
$res2 = mysql_command($sql2);
echo $sql2 . "<br/>";
$rowA = mysql_fetch_assoc($res2);
//echo "<strong>Alternative: </strong>" . $v. "<strong> Auto Alternative: </strong>" . $rowA['column10'] . "<br/>";
}
echo '</table>';
echo "<pre>";
print_r($my);
echo "</pre>";
在浏览器中,如果我回显查询和键:的结果是这样的:
SELECT column10 FROM `table` WHERE column1 = 'Villetta La Canoa'
SELECT column10 FROM `table` WHERE column1 = ' Casa Immerso nel Verde'
SELECT column10 FROM `table` WHERE column1 = ' La Rosetta'
SELECT column10 FROM `table` WHERE column1 = 'Agriturismo La Nonna'
SELECT column10 FROM `table` WHERE column1 = ' Villetta Cassiopeia'
SELECT column10 FROM `table` WHERE column1 = ' La Rosetta'
SELECT column10 FROM `table` WHERE column1 = 'Ca Gianca 2'
SELECT column10 FROM `table` WHERE column1 = ' Villetta Teresa'
SELECT column10 FROM `table` WHERE column1 = ' Appartamento Pinamare'
SELECT column10 FROM `table` WHERE column1 = ' Casa del Principe'
SELECT column10 FROM `table` WHERE column1 = 'Ca Gianca 2'
SELECT column10 FROM `table` WHERE column1 = ' Villetta Teresa'
SELECT column10 FROM `table` WHERE column1 = ' Appartamento Pinamare'
SELECT column10 FROM `table` WHERE column1 = ' Casa del Principe'
SELECT column10 FROM `table` WHERE column1 = 'Ca Gianca 2'
SELECT column10 FROM `table` WHERE column1 = ' Villetta Teresa'
SELECT column10 FROM `table` WHERE column1 = ' Appartamento Pinamare'
SELECT column10 FROM `table` WHERE column1 = ' Casa del Principe'
键,值为:
Key: 0 Value: Villetta La Canoa
Key: 1 Value: Casa Immerso nel Verde
Key: 2 Value: La Rosetta
Key: 3 Value: Agriturismo La Nonna
Key: 4 Value: Villetta Cassiopeia
Key: 5 Value: La Rosetta
Key: 6 Value: Ca Gianca 2
Key: 7 Value: Villetta Teresa
Key: 8 Value: Appartamento Pinamare
Key: 9 Value: Casa del Principe
Key: 10 Value: Ca Gianca 2
Key: 11 Value: Villetta Teresa
Key: 12 Value: Appartamento Pinamare
Key: 13 Value: Casa del Principe
Key: 14 Value: Ca Gianca 2
Key: 15 Value: Villetta Teresa
Key: 16 Value: Appartamento Pinamare
Key: 17 Value: Casa del Principe
我得到的是:
Alternative: Villetta La Canoa Auto Alternative: Villa Ronchi, Casa Ciserai, Villino Torretta, Casa Bianca
Alternative: Casa Immerso nel Verde Auto Alternative:
Alternative: La Rosetta Auto Alternative:
Alternative: Agriturismo La Nonna Auto Alternative: Agriturismo Antico Granaio, Casa Ciserai, Villa Ronchi, La Rosetta
Alternative: Villetta Cassiopeia Auto Alternative:
Alternative: La Rosetta Auto Alternative:
Alternative: Ca Gianca 2 Auto Alternative: Ca Gianca 1, La Vigna 2, Villetta Teresa
Alternative: Villetta Teresa Auto Alternative:
Alternative: Appartamento Pinamare Auto Alternative:
Alternative: Casa del Principe Auto Alternative:
Alternative: Ca Gianca 2 Auto Alternative: Ca Gianca 1, La Vigna 2, Villetta Teresa
Alternative: Villetta Teresa Auto Alternative:
Alternative: Appartamento Pinamare Auto Alternative:
Alternative: Casa del Principe Auto Alternative:
Alternative: Ca Gianca 2 Auto Alternative: Ca Gianca 1, La Vigna 2, Villetta Teresa
Alternative: Villetta Teresa Auto Alternative:
Alternative: Appartamento Pinamare Auto Alternative:
Alternative: Casa del Principe Auto Alternative:
谁能解释出什么问题?
从我的角度来看,您可能需要修剪$v以确保字符串的开头和结尾没有空格。
foreach($my as $k=> $v){
$v = trim($v);
此外,您应该尝试使用"LIKE"运算符而不是"="。也许在$v中添加"%"。
$sql2 = "SELECT column10 FROM `table` WHERE column1 LIKE '%{$v}%'";
mysql_command
到底是什么?我确定这是一个错字,你的意思是mysql_query
(唉,即使是这样:-/)
而且你不会得到一些结果,因为像这样的一些字符串后面有额外的空间
SELECT column10 FROM `table` WHERE column1 = ' Casa Immerso nel Verde'
^
您的示例数据证实了这一点。您别无选择的唯一结果是具有额外空间的结果。