Question

我有一个像这样的MySql表

Brand 
--------
Honda
Sozuki
Oddi
.
.
.

现在PHP中有一个动态数组：

$brand = ["Sozuki","Honda"]

问题是如何检索与上面数组中的单词匹配的数据。我试过这个：

$qry = 'select * from brand where brand Like '% $brand[0] %' and '% $brand[1] %';

它工作正常但如果我没有特定长度的数组，我会怎么做，你可以说动态数组。任何解决方案将不胜感激

Answer 1

你宁愿使用OR代替AND吗？

如果是这样，请使用类似的内容（注意： UNTESTED ）：

<?php
$brands     = array(1, 2, 3, 7, 8, 9);
$inQuery = implode(',', array_fill(0, count($brands), '?'));

$db = new PDO(...);
$stmt = $db->prepare(
    'SELECT *
     FROM brand
     WHERE brand IN(' . $inQuery . ')'
);

foreach ($brands as $k => $brand)
    $stmt->bindValue(($k+1), $brand);

$stmt->execute();

// Or simply $stmt->execute($brands);

Answer 2

你可以尝试：

$qry = 'select * from brand where brand Like ';
for ($i = 0; $i < sizeof($brands)-1; $i++) {
     $qry .= "'%" . $brand[$i] . "%'";
     if ($i < sizeof($brands) - 2) {
         $qry .= ' AND ';
     }
}

Answer 3

尝试

$qry = "select * from brand where brand ";
$i=1;
$count = count($brand); 
foreach($brand as $v) {
  $a = ($i < $count ? 'and' : '');
  $qry .=  " Like '% $v %'  $a ";
  $i++;
}
echo $qry;

Answer 4

<?php 
$i=0;
$brand=array("Volvo","BMW","Toyota","Honda","Suzuki");
$qry = 'select * from brand where brand Like '% $brand[i] `enter code here`%'';
$i++;
?>

Answer 5

您应该将数组定义为

$brand = array("Sozuki","Honda");

所以你可以使用$brand[0]等......

$qry = "SELECT * FROM brand WHERE brand LIKE '%{$brand[0]}%' AND '%{$brand[1])%'";

如何在SQL查询和PHP中匹配字符串中的单词并检索数据？

5 个答案: