php - 为什么准备语句会导致绑定(bind)中的参数数量错误？

Question

我得到了

mysqli_stmt_bind_param(): Number of variables doesn't match number of parameters 

警告

来自以下语句:

mysqli_stmt_bind_param($stmt, "ss", $str_addr_no, $str_addr);

我四处寻找 where ... like 示例，但看不出我的语法有什么问题。

$link = mysqli_connect("localhost","fred","password","h2o_amr");

$str_addr_no = $_REQUEST['meter_street_number'];
$str_addr    = $_REQUEST['meter_street_address'];

$stmt = mysqli_stmt_init($link);
mysqli_stmt_prepare($stmt, "select w.* from water w where w.mtr_addr_no like '?%' and w.mtr_addr_str like '?%' "))

mysqli_stmt_bind_param($stmt, "ss", $str_addr_no, $str_addr);

有没有办法让 PHP 显示或检查准备好的语句以查看实际准备了多少参数？

Answer 1

您引用了您的占位符:'?%'，这会将它们变成字符串，而不是占位符。占位符必须是 BARE ? 字符:

SELECT ... WHERE foo = ? // OK
SELECT ... WHERE Foo = '?' // no parameters here, just a string

对于通配符 LIKE 查询，您必须求助于 hack:

SELECT ... WHERE foo LIKE CONCAT('%', ?)