Php 自动递增变量

Php 自动递增变量,php,variables,Php,Variables,我试图使步骤数随着从数据库中提取的每个步骤而增加。这是我的密码: //get the project steps $ID=mysqli_real_escape_string($con,$_GET['id']); $projSteps="SELECT project_steps.projectStepDesc,project_steps.projectID FROM project_steps WHERE project_steps.projectID=$ID";

我试图使步骤数随着从数据库中提取的每个步骤而增加。这是我的密码:

//get the project steps 
$ID=mysqli_real_escape_string($con,$_GET['id']);         
      $projSteps="SELECT project_steps.projectStepDesc,project_steps.projectID FROM project_steps WHERE project_steps.projectID=$ID";  
    $results1 = mysqli_query($con,$projSteps);  
    if (!$results1) {
printf("Error: %s\n", mysqli_error($con));
exit();
}   
       if (mysqli_num_rows($results1) > 0){
        echo '<h2>Project Steps:</h2>';           

 while ($rows = mysqli_fetch_assoc($results1))
    {
      $a=1;
      echo "Step ". $a++ ."<br />";
      $psteps=$rows['projectStepDesc'];
      echo '<div id="steps">';
      echo "$psteps";
      echo '</div>';
      }}//end project steps
//获取项目步骤
$ID=mysqli\u real\u escape\u字符串($con,$\u GET['ID']);
$projSteps=“从project\u步骤中选择project\u步骤.projectID=$ID”中的project\u步骤.projectdesc,project\u步骤.projectID”;
$results1=mysqli_查询($con,$projSteps);
如果(!$results1){
printf(“错误:%s\n”,mysqli_错误($con));
退出();
}   
如果(mysqli_num_行($results1)>0){
回应“项目步骤:”;
而($rows=mysqli\u fetch\u assoc($results1))
{
$a=1;
回显“步骤”。$a++.“
”; $psteps=$rows['projectStepDesc']; 回声'; 回显“$psteps”; 回声'; }}//结束项目步骤

这只是在每个步骤上输出步骤1。我试着让它说第1步,第2步,第3步等等。我该怎么做呢?

您每次都在循环中设置
$a=1
。将其置于循环外部。

您正在while循环内部声明计数器。这将不起作用,因为,
$a=1。更改:

while ($rows = mysqli_fetch_assoc($results1))
    {
      $a=1;
致:


您已经将$a=1放在while之外,否则当循环运行时,每次a的值都会更改为1

    $ID=mysqli_real_escape_string($con,$_GET['id']);         
  $projSteps="SELECT project_steps.projectStepDesc,project_steps.projectID FROM         project_steps WHERE project_steps.projectID=$ID";  
$results1 = mysqli_query($con,$projSteps);  
if (!$results1) {
      printf("Error: %s\n", mysqli_error($con));
     exit();
   }   
   if (mysqli_num_rows($results1) > 0){
    echo '<h2>Project Steps:</h2>';           
     $a=1;
       while ($rows = mysqli_fetch_assoc($results1))
    {

  echo "Step ". $a++ ."<br />";
  $psteps=$rows['projectStepDesc'];
  echo '<div id="steps">';
  echo "$psteps";
  echo '</div>';
  }}//end project steps
$ID=mysqli\u real\u escape\u字符串($con,$\u GET['ID']);
$projSteps=“从project\u步骤中选择project\u步骤.projectID=$ID”中的project\u步骤.projectdesc,project\u步骤.projectID”;
$results1=mysqli_查询($con,$projSteps);
如果(!$results1){
printf(“错误:%s\n”,mysqli_错误($con));
退出();
}   
如果(mysqli_num_行($results1)>0){
回应“项目步骤:”;
$a=1;
而($rows=mysqli\u fetch\u assoc($results1))
{
回显“步骤”。$a++.“
”; $psteps=$rows['projectStepDesc']; 回声'; 回显“$psteps”; 回声'; }}//结束项目步骤
while
循环开始之前初始化
$a=1
。这是至关重要的,或者在每次迭代时将其重置为1;在外面,虽然这很好…不知道我怎么错过了…我会投票给这个答案。你能同意我的问题吗?当然。这是一件容易错过的事情,但从错误中吸取教训,你就会好起来。干杯
    $ID=mysqli_real_escape_string($con,$_GET['id']);         
  $projSteps="SELECT project_steps.projectStepDesc,project_steps.projectID FROM         project_steps WHERE project_steps.projectID=$ID";  
$results1 = mysqli_query($con,$projSteps);  
if (!$results1) {
      printf("Error: %s\n", mysqli_error($con));
     exit();
   }   
   if (mysqli_num_rows($results1) > 0){
    echo '<h2>Project Steps:</h2>';           
     $a=1;
       while ($rows = mysqli_fetch_assoc($results1))
    {

  echo "Step ". $a++ ."<br />";
  $psteps=$rows['projectStepDesc'];
  echo '<div id="steps">';
  echo "$psteps";
  echo '</div>';
  }}//end project steps