我将值保存为cookie,然后检查perfiles_vinculados
表中是否存在以获取perfil
表中具有相同ID的所有数据。
然后我创建一个$vinculado
结果的数组,并将其作为一行显示在HTML表格中。
问题是控制台返回:
可捕获的致命错误:
mysqli_result类的对象无法转换为字符串 C:\ XAMPP \ htdocs中\ miramonteapp \ API \ modal.php
Cookie:
document.cookie = "vinculaciones=" + $("#mod_id_perfil").val();
PHP:
//querys
<?php
include 'api/conexion.php';
$ides = $_COOKIE['vinculaciones'];
$juridicos = "SELECT perfil_juridica FROM perfiles_vinculados where perfil_fisica = '$ides'";
$con = mysqli_query($conexion, $juridicos);
$vinculado = mysqli_query($conexion, "SELECT * FROM perfil where id = '$con'");
?>
//table
<?php
while($reg = mysqli_fetch_array($vinculado)) {
$id = $reg['id'];
?>
<tr id="<?php echo " tr_ ".$reg['id']; ?>">
<td class="" data-id="<?php echo $reg['usuario'] ?>">
<?php echo $reg['nombre']; ?>
</td>
<td class="" data-id="<?php echo $reg['usuario'] ?>">
<?php echo $reg['cuit']; ?>
</td>
<td class="td-actions text-right">
<button type="button" rel="tooltip" class="btn btn-danger">
<i class="material-icons">close</i>
</button>
</td>
<?php } ?>
答案 0 :(得分:1)
您必须了解join
sql语句。
至于您当前的方法,首先需要从perfil_juridica
执行结果中获取$juridicos
值,然后将此值传递给第二个查询:
// first query
$juridicos = "SELECT perfil_juridica FROM perfiles_vinculados where perfil_fisica = '$ides'";
$result = mysqli_query($conexion, $juridicos);
$row = mysqli_fetch_array($result);
$perfil_juridica = $row['perfil_juridica'];
// second query
$vinculado = mysqli_query($conexion, "SELECT * FROM perfil where id = '$perfil_juridica'");
接下来应该做的是转到准备好的语句,而不是将不安全的值放入查询文本中。 This question会帮助你。