我正试图用两个表实现内部联接（Oracle称之为equixes），但我只是缺少了一些东西，因为它与MySQL内部联接不同。我也是PHP的新手（来自Lasso），所以为了自己的理智，我试图严格遵守我的语法。我已经看了关于这个错误的所有其他帖子，但它们都没有帮助。
使用PHP 5.5.29连接到Oracle 12c数据库，OCI8已在PHP信息页面中安装并验证。
Oracle对内部联接的解释
$login是纯文本，$pw在这个片段之前被散列：
// separate the login types by the "_" character
if (strpos($login, "_") == true) {
$fieldname = 'web_password_hash';
$dynfield = 'web_id';
$dynfield1 = 's.web_id';
} else {
$fieldname = 'student_web_password_hash';
$dynfield = 'student_web_id';
$dynfield1 = 's.student_web_id';
}
$schoolid = '(pcs.schoolid=4 OR pcs.schoolid=5)';
// search custom table for a field with the correct password (parent/student)
$dbc = oci_connect($psusername, $pspassword, $psconnection);
if(!$dbc) {
echo "Oracle Not connected";
} else {
$stmt = oci_parse($dbc, 'SELECT pcs.student_number AS snumber, :dynfield1 AS loginid, s.ID AS sid, s.family_ident AS famid, s.firstname AS fname, s.last_name AS lname, s.grade_level AS gradelevel
FROM pvsis_custom_students pcs, students s
WHERE pcs.student_number=s.student_number
AND :schoolid
AND pcs.field_name=:fieldname
AND pcs.string_value=:pw
AND s.enroll_status=0;');
// bind parameters to prevent SQL injection
oci_bind_by_name($stmt, ':dynfield1', $dynfield1);
oci_bind_by_name($stmt, ':fieldname', $fieldname);
oci_bind_by_name($stmt, ':schoolid', $schoolid);
oci_bind_by_name($stmt, ':pw', $pw);
// execute statement
oci_execute($stmt);
然后我得到的错误是：警告：oci_execute（）：ORA-00920：/path/to/file行XXX中的关系运算符无效
您不能像在这里那样绑定字段的名称。您需要动态构建查询：
$stmt = oci_parse($dbc, 'SELECT pcs.student_number AS snumber, ' . $dynfield1 . ' AS loginid, s.ID AS sid, s.family_ident AS famid, s.firstname AS fname, s.last_name AS lname, s.grade_level AS gradelevel
FROM pvsis_custom_students pcs, students s
WHERE pcs.student_number=s.student_number
AND ' . $schoolid . '
AND pcs.field_name=' . $fieldname . '
AND pcs.string_value=:pw
AND s.enroll_status=0;');
删除对:dynfield1、:fieldname和:schoolid的bind_by_name调用。
祝你好运。