php - PHP变量传递在数据库调用中不起作用

Question

这行得通。

function get_distinct_size_for_bracelets() {
    $sql = "SELECT DISTINCT size FROM mytable WHERE id = 27 AND type='plastic' ORDER BY size";
}

这不起作用并停止 php 死而没有错误报告。

 function get_distinct_size_for_bracelets($myvalue) {
    $sql = "SELECT DISTINCT size FROM mytable WHERE id = 27 AND type=".$myvalue." ORDER BY size";
}

我尝试了许多配置，但没有任何效果。

Answer 1

function get_distinct_size_for_bracelets($myvalue) {
    $sql = "SELECT DISTINCT size FROM mytable WHERE id = 27 AND type='".$myvalue."' ORDER BY size";
}

您仍然需要 SQL 查询中的单引号。

Answer 2

请记住引用传递的值：

function get_distinct_size_for_bracelets($myvalue) 
{ 
$sql = "SELECT DISTINCT size FROM mytable WHERE id = 27 AND type=".$myvalue." ORDER BY size";
}

应该：

function get_distinct_size_for_bracelets($myvalue) 
{ 
$sql = "SELECT DISTINCT size FROM mytable WHERE id = 27 AND type='".$myvalue."' ORDER BY size";
}

请注意添加的单引号type。

Answer 3

你仍然需要单引号。所以

$sql = "SELECT DISTINCT size FROM mytable WHERE id = 27 AND type='".$myvalue."' ORDER BY size";

Answer 4

你没有逃避你的价值，你忘记了你的单引号，那是我的猜测。尝试：

function get_distinct_size_for_bracelets($myvalue) { 
    $query = sprintf("SELECT DISTINCT size FROM mytable WHERE id = 27 AND type='%s'  ORDER BY size",
        mysql_real_escape_string($myvalue));
}

这使您可以将转义值传递到字符串中，而不是使用连接。

Answer 5

尝试

function get_distinct_size_for_bracelets($myvalue) {
    $sql = "SELECT DISTINCT size FROM mytable WHERE id = 27 AND type='".$myvalue."' ORDER BY size";
}

Answer 6

MySQL 也有不同的数据类型。字符串也需要用引号引起来：

$sql = "SELECT DISTINCT size FROM mytable WHERE id = 27 AND type='".$myvalue."' ORDER BY size";

或者更好地使用该mysql_real_escape_string功能：

$sql = "SELECT DISTINCT size FROM mytable WHERE id = 27 AND type='".mysql_real_escape_string($myvalue)."' ORDER BY size";