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修正model类join方法不能读取指定表前缀的问题

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oldrind
2016-03-30 10:11:09 +08:00
parent 9ca3a29b59
commit 1ef37d253d
1 changed files with 7 additions and 9 deletions
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16
library/think/Model.php
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@@ -1477,17 +1477,15 @@ class Model
$prefix = $this->tablePrefix;
// 传入的表名为数组
if (is_array($join)) {
if (0 !== key($join)) {
// 键名为表名，键值作为表的别名
$table = key($join) . ' ' . current($join);
if (0 !== $key = key($join)) {
// 设置了键名则键名为表名，键值作为表的别名
$table = $key . ' ' . array_shift($join);
} else {
$table = current($join);
$table = array_shift($join);
}
if (isset($join[1])) {
// 第二个元素为字符串则把第二元素作为表前缀
if (is_string($join[1])) {
$table = $join[1] . $table;
}
if (count($join)) {
// 有设置第二个元素则把第二元素作为表前缀
$table = (string) current($join) . $table;
} else {
// 加上默认的表前缀
$table = $prefix . $table;