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Do not display error for foreign keys to other databases

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This commit is contained in:
Jakub Vrana 2019-08-21 14:19:48 +02:00
parent 021e3f900d
commit f37a878b62
1 changed files with 8 additions and 7 deletions
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15
adminer/foreign.inc.php
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@ -48,17 +48,18 @@ if ($_POST) {
$row["table"] = $TABLE; $row["table"] = $TABLE;
$row["source"] = array(""); $row["source"] = array("");
} }
$source = array_keys(fields($TABLE)); //! no text and blob
$target = ($TABLE === $row["table"] ? $source : array_keys(fields($row["table"])));
$referencable = array_keys(array_filter(table_status('', true), 'fk_support'));
?> ?>
<form action="" method="post"> <form action="" method="post">
<p> <p>
<?php if ($row["db"] == "" && $row["ns"] == "") { ?> <?php
<?php echo lang('Target table'); ?>: if ($row["db"] == "" && $row["ns"] == "") {
<?php echo html_select("table", $referencable, $row["table"], "this.form['change-js'].value = '1'; this.form.submit();"); ?> $source = array_keys(fields($TABLE)); //! no text and blob
$target = ($TABLE === $row["table"] ? $source : array_keys(fields($row["table"])));
$referencable = array_keys(array_filter(table_status('', true), 'fk_support'));
echo lang('Target table') . ": ";
echo html_select("table", $referencable, $row["table"], "this.form['change-js'].value = '1'; this.form.submit();");
?>
<input type="hidden" name="change-js" value=""> <input type="hidden" name="change-js" value="">
<noscript><p><input type="submit" name="change" value="<?php echo lang('Change'); ?>"></noscript> <noscript><p><input type="submit" name="change" value="<?php echo lang('Change'); ?>"></noscript>
<table cellspacing="0"> <table cellspacing="0">