Php 如何检查字段是否为加载表中的外键?
下面是我的脚本,用于加载表格并将其显示在HTMLPhp 如何检查字段是否为加载表中的外键?,php,sql,codeigniter,postgresql,Php,Sql,Codeigniter,Postgresql,下面是我的脚本,用于加载表格并将其显示在HTML表格中 $fields = $this->db->list_fields($tableName); $fieldData = $this->db->field_data($tableName); $this->db->from($tableName); $this->db->order_by($fields[0], "asc"); $query = $this->db->get();
表格中
$fields = $this->db->list_fields($tableName);
$fieldData = $this->db->field_data($tableName);
$this->db->from($tableName);
$this->db->order_by($fields[0], "asc");
$query = $this->db->get();
$table = $query->result();
$columnsQ = $this->db->query("select count(column_name) from information_schema.columns where table_name = '".$tableName."'");
$columns = $columnsQ->row();
foreach($table as $row) {
echo "\t<tr class='value'>\n";
$colNr = 1;
foreach ($row as $tdVal) {
$colNr0 = $colNr-1;
$valType = $fieldData[$colNr0]->type;
$isForeign; // how to get info?
if($colNr==1) {
echo "\t\t<td data-valType='$valType' class='id'>$tdVal</td>\n"; }
else {
echo "\t\t<td data-isForeign='$isForeign' data-valType='$valType' data-colNr='$colNr'>$tdVal</td>\n"; }
$colNr++;
}
}
</table>";
$fields=$this->db->list\u字段($tableName);
$fieldData=$this->db->field\u data($tableName);
$this->db->from($tableName);
$this->db->order_by($fields[0],“asc”);
$query=$this->db->get();
$table=$query->result();
$columnsQ=$this->db->query(“从信息_schema.columns中选择count(column_name),其中table_name='”“$tableName.”);
$columns=$columnsQ->row();
foreach($行的表格){
回显“\t\n”;
$colNr=1;
foreach(行作为$tdVal){
$colNr0=$colNr-1;
$valType=$fieldData[$colNr0]->type;
$isForeign;//如何获取信息?
如果($colNr==1){
echo“\t\t$tdVal\n”}
否则{
echo“\t\t$tdVal\n”}
$colNr++;
}
}
";
在$isForeign
下应该有什么查询来获取加载值是否为外键的信息?看看这个问题的答案,您只需要在迭代foreach时为特定列添加所需的额外约束
我正在努力找出这个问题中的不同之处