WordPress或php奇数语法错误和google chrome ajax登录
所以我在搜索表,我有1500行代码,它工作得很好,但我有两个问题,一个是我的代码错误,如下所示:WordPress或php奇数语法错误和google chrome ajax登录,php,ajax,wordpress,google-chrome,Php,Ajax,Wordpress,Google Chrome,所以我在搜索表,我有1500行代码,它工作得很好,但我有两个问题,一个是我的代码错误,如下所示: if ($wpdb->get_var('show tables like "' . placeholders . '"') == placeholders) { foreach ($placeholders as $placeholder_) { $Gplaceholders[$placeholder_] = $wpdb->get_row('SELECT * FR
if ($wpdb->get_var('show tables like "' . placeholders . '"') == placeholders) {
foreach ($placeholders as $placeholder_) {
$Gplaceholders[$placeholder_] = $wpdb->get_row('SELECT * FROM `' . placeholders . '` WHERE `placeholder` = "' . $placeholder_ . '"', ARRAY_A)['value'];
}
}
错误:
Parse error: syntax error, unexpected [ in ******
另一个是我的ajax登录和注册在Mozilla上运行得非常好,但在chrome上它只是截击而不做任何事情,有人在chrome中遇到过这种问题吗
$Gplaceholders[$placeholder_] = $wpdb->get_row('SELECT * FROM `' . placeholders . '` WHERE `placeholder` = "' . $placeholder_ . '"', ARRAY_A)['value'];
您的第一个占位符不需要用“…”包装尝试:
$Gplaceholders[$placeholder_] = $wpdb->get_row('SELECT * FROM `placeholders` WHERE `placeholder` = "' . $placeholder_ . '"', ARRAY_A)['value'];
您确定要在“$Gplaceholders”中添加“G”吗?可能是PHP版本不支持数组解引用的版本问题,即<5.4.0