Question

我目前正在使用一个表格,使用<条码>how栏,以便把所有栏目纳入核对箱。下面的法典就是这样做的。我试图在问询中设定另一个查询权,以寻找一个有相同栏目的不同表格,如果表中的行文有0或1个显示“检查”箱。之所以使用“单列”查询,是因为各项目总是增加一栏。

/////EDIT after posting I realized that it will be better just to use the same table since the "Columns" represent all the "checkboxes" when output anyway. But the same problem still exists. I am trying to double up on the queries. Query 1 = Extract all columns and create checkboxes Query 2 = use row that matches $newloc and if checkboxes are in specific columns mark out put columns with "Checked" boxes. I have another query on the same page that populated user information which works fine. This page works as a confirmation page which has all user information as well as certain items they have (the Checkboxes) and then they can edit on this page and submit if needed. I want to be clear on all that is in this page just in case that effects this part. One other thing I need to add is since I am using the same table I need to skip the first 3 columns since that is other data that I dont want as columns.

 <?php 
  $conc = new mysqli("127.0.0.1:3306", "root", "", "mydb");
  $res= $conc ->query("SHOW COLUMNS from matmemmatrix");
  
  // this part here is what I was imagining would work
  $sql = "SELECT * FROM matmemmatrix WHERE memid1 =  $uid  AND memlocid =  $newloc ";
  $datam = mysqli_query($conc, $sql);
  $a = 1;
  $row1 = $datam[$a]; // first column ?? 
  //      


  echo "<html>";
  echo "<body>";               
  echo "<table>";
  $i = 0;
  while ($row = $res->fetch_assoc()) {
     $id = $row[ Field ]; 


       // EDIT : check to see if $id matches $row1
     if($id = $row1){
        $checked = 1;
        }else{
        $checked = 0;
        }
        //



     if($i > 5 ){
        $i== 0 ;
    
     if($i== 5 ){
        echo "</tr>";
        echo "<tr>";
        $i = 0 ; 
     ;   
     if ($i == 0  ){
        echo "<tr>";
  };
  echo "<td>";
  echo  <input type ="checkbox"  
                       value=" .htmlspecialchars($id). "  
                           id=" .htmlspecialchars($id). "          
                          name=" .htmlspecialchars( mat[] ). "  
                        checked=  .$checked. >
                        </option> ;
 echo  <label for =" .htmlspecialchars($id). "> .htmlspecialchars($id). </label> ;
 echo "</td>";
 $i++;
 // EDIT Increment next column to see if 0 or 1
 $a++;
//
 }
  echo "</table>";
  echo "</body>";
  echo "</html>";   
?>

我试图利用下文中的询问,以获得“检查”栏。但失败。

$sql = "SELECT * FROM matmemmatrix WHERE memid1 =  $uid  && memlocid =  $newloc ";

I couldn t get any results from that query to work with the above query.

我试图利用上述询问本身,提取所有栏目作为检查箱使用,并使用1或0的浏览数据作为检查但失败的标志。

能否从这个栏目中提取一栏作为检查箱,并利用浏览数据显示,如果1或0,将进行核对? 我想确保表格中的所有栏目都表明它们是否受到检查。

Answer 1

I got this code to work from 2 queries. Needs a little cleaning though.

<?php
$conc = new mysqli("127.0.0.1:3306", "root", "", "database");           
$sqla = "SHOW COLUMNS from matmemmatrix";
$res= mysqli_query($conc, $sqla);
$uid = $_SESSION[  uid  ];
$newloc = $_SESSION[  newloc ];

$sql = "SELECT * FROM matmemmatrix WHERE memid1 =  $uid  AND memlocid =  $newloc ";
$resu = $conc->query($sql);
$datam = $resu->fetch_assoc();
$a = "0";
echo "<html>";
echo "<body>";               
echo "<table>";             
$i = 0;
while ($row = $res->fetch_assoc()) {
$id = $row[ Field ]; //prints column names
$id1 = $rowa[ MatName ]; 
if ($datam[array_keys($datam)[$a]] == "1")
{ $checked = "checked";
}else{
$checked = "";
}
$a++;
if($i > 5 ){
$i== 0 ;
};

if($i== 5 ){
echo "</tr>";
echo "<tr>";      
$i = 0 ;           
};
if ($i == 0  ){
echo "<tr>";
};
echo "<td>";
echo  <input type ="checkbox"  value=" .htmlspecialchars($id). "  id=" .htmlspecialchars($id). "  name=" .htmlspecialchars( mat[] ). "   .$checked. >
</option> ;
echo  <label for =" .htmlspecialchars($id). "> .htmlspecialchars($id). </label> ;
echo "</td>";
$i++;
                  
}   
echo "</table>";
echo "</body>";
echo "</html>";   
$_POST[ access_listb ];
?>

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