使用Array - PHP创建一个动态MySQL查询 - Create a Dynamic MySQL Query using an Array - PHP

我正在尝试运行MySQL查询，其中一块正在使用LIKE查找。

表结构:TableExample

id    name     piece   
0     jon      piece0
1     james    piece3
2     sarah    piece6

到目前为止的剪辑:

$pieces  = "piece0 piece1";  //variable
$piecearrayexplode = explode(" ", $pieces);
$piece0 = $piecearrayexplode[0];
$piece1 = $piecearrayexplode[1];
$sql = "SELECT * FROM TableExample WHERE piece LIKE '%$piece0%' OR pieces LIKE '%$piece1%'";

我的问题是$pieces是一个变量，我需要$sql是动态的，并自动提供正确数量的LIKE语句。

。如果$pieces = "piece0 piece1 piece2"，我希望$sql是:

$sql = "SELECT * FROM TableExample WHERE piece LIKE '%$piecearrayexplode[0]%' OR pieces LIKE '%$piecearrayexplode[1]%' OR pieces LIKE '%$piecearrayexplode[2]%'";

注意:$pieces总是用空格分隔。

我可以做字数统计。

$count = str_word_count($pieces);

我不知道该怎么办。

使用php变量创建一个动态mysql查询

这似乎不是我要找的，因为like是连续的，而不是像WHERE这样的单个语句。我遗漏了什么吗?

所以动态构建你的查询:

$foo = '... list of pieces ...';
$parts = explode(' ', $foo);
$likes = array();
foreach($parts as $part) {
   $likes[] = "piece LIKE '%$part%'";
}
$sql = "SELECT ... WHERE " . implode(' or ', $likes);

您可以迭代throw数组并逐个添加片段进行查询。(未测试)

$pieces = "piece0 piece1 piece2";
$piecesArr = explode(" ", $pieces);
$sql = "SELECT * FROM TableExample WHERE piece LIKE"; 
$first = true;
foreach ( $piecesArr as $PA ) {
   if ( $first ) {
     $sql .= " '%$PA%'";
     $first = !$first;
   }
   else $sql .= " OR pieces LIKE '%$PA%'"`;
}

我所做的基本上是从视图动态生成我的结果。然后，我在视图中填充具有列名值的下拉列表，然后将这些变量从视图传递给select语句。

<?php
error_reporting(0);
# FileName="Connection_php_mysql.htm"
# Type="MYSQL"
# HTTP="true"
$hostname_connection = "localhost";
$database_connection = "xxx";
$username_connection = "root";
$password_connection = "";
$connect = mysql_pconnect($hostname_connection, $username_connection, $password_connection) or trigger_error(mysql_error(),E_USER_ERROR);
mysql_select_db($database_connection,$connect);
$col1=$_POST['col1'];
$col2=$_POST['col2'];
$col3=$_POST['col3'];
$result = mysql_query("SELECT $col1 as column1,$col2 as column2,$col3 as colmun3 FROM variety_view");
if(!$result){
echo "failed";
} else {
echo "Perfecto";
}
echo "<table>"; // start a table tag in the HTML
while($row = mysql_fetch_array($result)){   //Creates a loop to loop through results
echo "<tr>
<td>" . $row['column1'] . "</td>
<td>" . $row['column2'] . "</td>
<td>" . $row['colmun3'] . "</td>
</tr>";  //$row['index'] the index here is a field name
}
echo "</table>"; //Close the table in HTML
?>
<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">
<html xmlns="http://www.w3.org/1999/xhtml">
<head>
<meta http-equiv="Content-Type" content="text/html; charset=iso-8859-1" />
<title>Untitled Document</title>
</head>
<body>
<form method="Post" action="">
<select name="col1">
<option value="official_name">Variety Name</option>
<option value="localname">Local name</option>
<option value="country">Country</option>
<option value="pedigree">Pedigree</option>
</select> &nbsp;
<select name="col2">
<option value="official_name">Variety Name</option>
<option value="localname">Local name</option>
<option value="country">Country</option>
<option value="pedigree">Pedigree</option>
</select> &nbsp; &nbsp;
<select name="col3">
<option value="official_name">Variety Name</option>
<option value="localname">Local name</option>
<option value="country">Country</option>
<option value="pedigree">Pedigree</option>
</select> <br />
<input type="submit" value="Search" />
</form>
</body>
</html>